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Calculus 
DEL GRANDE/DUFF 
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Elements of Modern Mathematics 
Calculus 
SECOND EDITION 
). I. DEL GRANDE, M.A. 
G. F. D. DUFF, M.A., Ph.D., F.R.S.C. 
EDUCATIONAL PUBLISHING COMPANY 
A DIVISION OF CANADA PUBLISHING CORPORATION 
TORONTO ONTARIO CANADA 
COPYRIGHT ©GAGE PUBLISHING LIMITED, 1979 
PRINTED AND BOUND IN CANADA 
All rights reserved. No part of this book may be reproduced in any 
form without permission in writing from the publisher. Reproducing 
passages from this book by mimeographing or by photographic, 
electrostatic or mechanical means without the written permission of 
the publisher is an infringement of copyright law. 
ISBN 0-7715-3634-8 
9 10 11 12 BP 88 87 86 
METRIC COMMISSION HAS GRANTED USE OF 
THE NATIONAL SYMBOL FOR METRIC CONVERSION 
CONTENTS 
/Introduction to Calculus 
1.1 Area 2 
1.2 Slope of the Tangent to a Curve 6 
1.3 Limits 10 
u Limit of a Sequence 16 
1.5 Sums of Infinite Series 21 
1.6 Special Series 27 
1.7 Calculation of Areas 33 
1.8 Properties of Limits 39 
1.9 Review Exercise 42 
2 /Slopes and Rates of Change 
2.1 Slope of a Linear Function 46 
2.2 Slope of the Tangent Line 50 
2.3 Tangents to a Curve 55 
24 The Derivative Function 60 
2.5 Rates of Change 66 
2.6 Application to Curve Sketching 71 
2.7 Continuous Functions 74 
2.8 Review Exercise 79 
iii 
3/ 
3.1 
Derivatives and their Properties 
The Derivative as a Limit 
4! 
83 
3.2 The Derivative of a Power 88 
3.3 Derivative of a Sum 94 
34 Derivative of a Product 98 
3.5 Derivative of a Quotient 102 
3.6 The Chain Rule 106 
3.7 The Tangent Lnie 111 
3.8 Tangents to More General Curves 112 
3.9 A Special Trigonometric Limit 117 
3.10 Derivative of the Sine Function 121 
3.11 Further Trigonometric Derivatives 124 
3.12 Derivative of an Inverse Function 129 
3.13 From Psychology to Logarithms 133 
3.14 The Exponential Function 141 
3.15 Growth ayid Decay 142 
3.16 Review Exercise 144 
Applications of Derivatives — Motion 
? 
U Velocity —Average and Instantaneous 148 
4.2 The Calculation of Velocities 154 
4.3 Acceleration 160 
44 Acceleration Problems 165 
4.5 Calculation of Related Rates —Areas and Volumes 172 
4.6 Rate Problems Involving Distances 179 
4.7 Review Exercise 184 
IV 
Applications of Derivatives — 
Qualitative Properties of Graphs 
5.1 Increasing and Decreasing Functions 189 
5.2 Maximum and Minimum Values 195 
5.3 Maximum and Minimum Problems 201 
54 The Best A ngle 209 
5.5 The Second Derivative 211 
5.6 Hills and Valleys 216 
5.7 Sufficient Conditions for Maximum or 
Minimum Values 223 
5.8 Review Exercise 230 
Functions with Given Derivatives 
6.1 Functions with Derivative ax + b 236 
6.2 Functions with Derivative a Power 241 
6.3 Families of Curves with Given Slope 247 
64 The Differential Equation = fix) 
rf'Y 
252 
6.5 
\A/%Kj 
Motion with Given Velocity 258 
6.6 Motion with Given Acceleration 265 
6.7 Differentials 271 
6.8 Applications to Business and Population Problems 275 
6.9 Review Exercise 280 
v 
Area and Integration 
7.1 Area Functions 287 
7.2 The Rate of Change Relation for Areas 292 
7.3 An Equation for the Area Function 297 
74 Areas Between Two Curves 302 
7.5 hide finite Integration 307 
7.6 Indefinite Integral —Evaluation by Substitution 310 
7.7 Further Methods of Indefinite Integration 315 
7.8 Integration by Parts 322 
7.9 Review Exercise 327 
Further Applications of Integration 
8.1 Areas and Definite Integrals 330 
8.2 Integration as a Process of Summation 333 
8.3 Areas 335 
84 Volumes 339 
8.5 Volumes of Revolution 345 
8.6 Work arid Pressure 348 
8.7 Length of a Curve 354 
8.8 Areas of Surfaces of Revolution 358 
8.9 Buff on’s Needle Problem 362 
8.10 Average Value of a Function 365 
8.11 Averages and Population Distributions 372 
8.12 Review Exercise 377 
vi 
0 /Approximations of Functions by Series 
9.1 Functions and Polynomials 380 
9.2 Power Series 387 
9.3 Madmirin’'s Series 390 
94 The Exponential Series 394 
9.5 The Logarithmic Series 397 
9.6 Series for sin x and cos x 401 
9.7 Series Expansion of I ntegrals 405 
9.8 Review Exercise 408 
10/ Polar Coordinates 
10.1 The Polar Coordinate System 411 
10.2 Polar Graphs 414 
10.3 Areas in Polar Coordinates 417 
104 Arc Length in Polar Coordinates 420 
10.5 Complex Numbers avid Polar Coordinates 423 
10.6 Review Exercise 431 
vii 
Tables of Derivatives and Integrals 
Numerical Tables 
Answers 
433 
439 
449 
Index 485 
PREFACE TO THE SECOND EDITION 
Because of the unique importance of the basic concepts of cal¬ 
culus, and because of its many applications, calculus has long been 
the cornerstone of higher mathematical analysis. In calculus the 
student learns how to apply his knowledge of functions in many 
interesting new situations involving limits, rates of change and 
integration processes. 
Calculus is designed for a full-year introductory course. After an 
introductory chapter describing the two basic problems of calculus 
and their common relationship to the concept of limit, the differen¬ 
tial calculus is treated in some detail. As well as stressing clarity of 
basic concepts, the exposition emphasizes qualitative relationships 
involving rates of change and inequalities, and a diversity of applica¬ 
tions including modern topical problems. 
In the treatment of integral calculus, the methods of anti¬ 
differentiation and of summation leading to definite integration are 
developed in a number of stages, and eventually are brought to¬ 
gether and compared. The technique of indefinite integration is 
developed with attention to algebraic pattern and motivation. The 
applications of integration are designed to cover a wide range for an 
introductory course, and some selection is possible if time is limited. 
This second edition of Calculus has been metricated, updated and 
revised. The suggestions of several highly respected teachers result¬ 
ed in many changes and improvements. About 300 new problems 
have been added in strategic places. The introduction to logarithms 
is motivated by an interesting psychological property. There are 
new sections on growth and decay, best angle, and applications to 
business and population. A probability model that leads to the 
value of 7r is used to solve the Buffon needle problem. 
The first edition of Calculus also appears in Spanish under the 
title of Introduccion al calculo elemental. This second edition appears 
in a French version, making Calculus appropriate for all Canadian 
classrooms. 
IX 
A NOTE TO THE STUDENT* 
The differential and integral calculus is a very interesting and 
rewarding subject at the threshold of modern mathematics. You 
will learn several new, powerful, and intriguing concepts that turn 
out to be very widely useful and remarkably effective in opening 
new vistas. Calculus is a new and higher mathematical synthesis 
involving most of the ideas you have already encountered. Practise 
looking for applications in the world around you. 
In order that you should have the best opportunity to master the 
work of this course, we therefore recommend: 
(1) You should work hard and steadily at the material, including 
examples and exercises, from the very first week of the course. If at 
the outset you fall behind, you will have the greatest difficulty in 
overcoming the eventual effects of this. 
(2) Do not be discouraged if some difficulties prove to be persistent 
or take some time to be resolved. The very best mathematicians 
struggled with calculus for more than a hundred years before an 
adequate definition of basic concepts such as limit was found. You 
will gain much more from a greater mastery of the basic ideas if you 
exercise patience and resolution. 
(3) When the course is well under way, initiatea running review of 
the work two or three chapters earlier, and rework a few exercises 
in detail. Note the proportion of earlier difficulties that have now 
disappeared. If some remain, review again later. 
(4) Pay attention to workmanship. Neatness and clarity of expres¬ 
sion are potent allies, well worth an effort of cultivation. Correct 
mathematical grammar (use of equality signs) and logic (as in 
necessary vs. sufficient conditions) are important. Always draw a 
large clear diagram. 
^Written after the authors had reviewed the results of marking 200 
Calculus papers. 
x 
Introduction to Calculus 
Throughout the history of mathematics problems involving the con¬ 
cept of a function have been encountered. Archimedes (287-212 
B.C.) calculated certain areas and volumes, such as the volume of 
a sphere, by methods that were far ahead of their time. This long 
evolution of the mathematics of functions took its greatest step 
forward in the 17th century when it was gradually realized that 
certain useful, new operations can be defined using functions. In 
striving to continue the works of Kepler and Galileo, the spirited 
scholars of 17th century Europe were led to consider two main 
problems involving graphs or curves, and therefore functions. 
The first problem was to find the area of a region with given 
curved boundaries, while the second was to construct the tangent 
at a given point of a curve. 
Credit for the decisive steps in solving these problems, and dis¬ 
covering calculus, goes to Newton and Leibniz. They established a 
connection between thf two problems, and showed how they could 
be resolved by means of new operations, which are now called inte¬ 
gration and differentiation and which are performed using functions. 
Their methods of calculation, which became popularly known as 
calculus, form one of the most exciting and powerful achievements 
of modern mathematics. 
Throughout the 18th and 19th centuries the differential and 
integral calculus was extensively developed by many gifted math¬ 
ematicians including Euler, Lagrange, Gauss, and Cauchy, and it 
also became a cornerstone of modern physical science. Within 
1 jIntroduction to Calculus 1 
mathematics, calculus has given rise to analysis, one of the main¬ 
streams of modern mathematics. In recent times, many other dis¬ 
ciplines of pure and applied science, commerce and industry have 
increasingly emphasized cjuantitative and mathematical approaches. 
Thus the differential and integral calculus is remarkable not only 
for its own power and fascination but also for the great role it has 
played in the evolution of science and thought since it was invented 
three hundred years ago. 
1.1 Area 
Two hundred years before Archimedes, the mathematician 
Antiphon attempted to find the area of a circle by drawing inscribed 
and circumscribed regular polygons in and about the circle. His 
method is called the “process of exhaustion” because as the number 
of sides of the polygon is increased, the area between the polygons 
was ultimately exhausted. See Figure 1.1. 
Archimedes, in his time, proved that the area of a circle was equal 
to ^ circumference X radius. To find the circumference he noted 
that its length must he between the respective perimeters of the 
inscribed and circumscribed polygons of Figure 1.1. As the number 
of sides of the regular polygons is increased, the two perimeters 
approach each other in value. The circumference always remains 
between the two values. 
For a circle of diameter 1, the following results are obtained. 
No. of 
Sides 
Perimeter of 
Inscribed Polygon 
Perimeter of 
Circumscribed Polygon 
Difference 
4 2.828 4.000 1.172 
8 3.061 3.314 0.253 
12 3.106 3.214 0.109 
24 3.133 3.160 0.027 
36 3.138 3.150 0.012 
9 1.1 /Area 
If a circle has diameter 1, its radius is \ and its circumference, 
2wr, is 7r. Notice how the two perimeter values approach each other 
— and the value of t is always between them. 
Archimedes used regular polygons of 96 sides in order to get his 
approximation for 7r. 
Archimedes also found the area of an ellipse and the area of a 
segment of a parabola cut off by any chord, using a special technique. 
We shall demonstrate Archimedes’ technique by finding the area 
of the region bounded by y = x2 + 1 and the lines x = 0, x = 3, 
and y = 0. 
y 
A = 1+2+5 
= 8 
A =2+5 + 10 
The shaded area under the parabola must lie between the areas 
of rectangle DOCE and rectangle AOCB\ that is, between 3 and 30 
square units. For a better estimation of the area we try to “squeeze” 
its value between two values which can be made as close together 
as we like. 
The horizontal interval 0 to 3 is divided into 3 equal parts and 
two sets of rectangles are drawn, one set with the left hand corner 
of each rectangle on the parabola, and the second set with the right 
hand corner on the parabola. The sums of the areas of the rectangles 
are respectively 8 and 17 square units. The area, A, of the required 
region lies between these two numbers. 
Thus, 8 < A <17. 
1.1/Area 3 
We now divide the interval from 0 to 3 into six equal parts and 
construct two sets of rectangles as before but each having a width 
of \ unit. 
We wdl give you the opportunity of completing the calculations 
to show how the area of the region is being “squeezed" between two 
number sequences. The two number sequences are said to approach 
the area value, and the process is called finding a limit. 
The idea of the method of limits is very simple. In order to find the 
exact value of some magnitude, we first determine not the magnitude 
itself, but some approximation to it. However, we make not just 
one approximation but a whole sequence of them, each more ac¬ 
curate than the last. Then by examining this chain of approxima¬ 
tions, we are able to determine its limit, which is the exact value of 
the magnitude. 
Exercise 1.1 
Some questions in this exercise lend themselves to computer 
solutions. 
O 
1. hind the areas of the inscribed and circumscribed squares of a 
circle with radius 1. 
4 
1.1 / Area 
2. Reconstruct Antiphon's calculations using the steps that fol¬ 
low for a regular inscribed and circumscribed hexagon. 
(a) How many congruent sides has a regular hexagon? 
(b) Join each vertex to the centre of the circle. 
(c) How many congruent triangles are formed? 
(d) Calculate the angle measure at the centre for each triangle. 
(e) Find the area of each triangle using the formulas given. 
, , 0 
A = a~ tan — 
j-j 
(/) Find the area of the inscribed hexagon. 
(g) Find the area of the circumscribed hexagon. 
3. Repeat Question 2 using regular 8-gons, 12-gons, 24-gons, and 
36-gons. (Archimedes used 96-gons for his approximation of w. 
The full translation of Archimedes’ elaborate calculation can 
be found on pages 50-55 in Sir Thomas Heath’s History of 
Greek Mathematics, Vol II.) 
4. Use the results of the previous questions to complete the follow¬ 
ing table. 
n 4 6 8 12 24 36 
Inscribed area 
Circumscribed area 
1.1/Area 5 
Prove. 
6. Find the perimeter of a 60-sided regular polygon inscribed in a 
circle of diameter 1. 
7. Find the perimeter of a 60-sided regular polygon circumscribed 
about a circle of diameter 1. 
8. Find the sum of the areas of the two sets of rectangles in the 
problem of finding the area of the region bounded by y = x2 + 1, 
x = 0, y = 0 and x = 3 for each of the following cases. 
The width of each rectangle is 
The width of each rectangle is -j. 
The width of each rectangle is 3^. 
Write a computer program to output the sum of the areas 
of each set of rectangles where the interval is 0.01 units. 
Find the sum of the areas of the two sets of rectangles that 
approximate the area in the first quadrant bounded by the 
X2 y2 
ellipse — -p — = 1, the lines x = 0 and y = 0, where the inter- 
zo 10 
val is 0.01 units. 
1.2 Slope of the Tangentto a Curve 
We know how to find the slope of a line 
from its defining equation. For example, 
the slope of the line that is defined by 
3x + 4y — 12 = 0 is —f. The slope of the 
tangent to a circle is also easy to determine 
because we know that the tangent line is 
always perpendicular to the radius drawn 
to the point of contact. 
What do we mean by a tangent to a non-linear curve? Newton and 
Leibniz applied their new technique to the problem. 
6 1.2/Slope of the Tangent to a Curve 
Example 1 
Find the slope of the tangent to the parabola y = x2 + 1 at P(2, 5). 
Solution 
Take any other point nearby to the right of P(2, 5) on the curve. 
Call it <2(2 + h, (2 + h)2 + 1) where h > 0. The slope of secant PQ is 
(2-M)> + i-6_4>+j;_ m 
2 + & - 2 h 
Now consider the point i? to the left of P{2, 5) with coordinates 
(2 — h, (2 — /j)2 + 1), where again /? > 0. 
The slope of the secant PR is 
(2 - h)2 + 1 - 5 -4h + h2 
h — 2 ■ h 
4 — h (h 9^ 0). 
Notice that the slopes of all secants 
P<2 are greater than 4 while the slopes 
of all secants PR are less than 4. Each 
secant in either set intersects the 
curve at two points. We have located 
or “sandwiched” the value of the 
slope of the tangent between two 
numbers. 
Thus, 4 — h < tangent slope < 4 + h. 
1.2/Slope of the Tangent to a Curve 
As Q and R approach P along the curve, the value of h approaches 
zero, and the tangent slope is “squeezed” between two numbers each 
approaching 4. Consequently the tangent slope can only be 4. 
Notice in particular that 4 + h approaches 4 as h approaches 
zero. We say that the limit of 4 + h is 4 as h approaches zero and 
we write 
lim (4 + h) = 4. 
h—>0 
Notice also that if there had been a break or a corner in the curve 
at (2, 5), our method would have failed, because the two values that 
“sandwich” the tangent slope would not have approached the same 
number. For this reason we insist that the curve be smooth at the 
point of tangency. No breaks or corners are allowed. 
O 
x 
O 
x 
Exercise 1.2 
1. Use Archimedes’ method for finding the slope of the tangent at 
the given point for each of the following curves. 
(a) y = 2x* 1 2 at (3, 18) 
(b) y = 3x2 + 1 at (2, 13) 
(c) y = x3 at ( — 1, — 1) 
2. (a) Graph the function de¬ 
fined by y = -, x > 0. 
x 
ib) Draw tangents at (^,2), 
(1, 1), and (2, ^). 
(c) Find the slope of the tan¬ 
gents by measurement. 
(d) Use Archimedes’ method 
to calculate the slope of 
each tangent. 
y 
8 1.2/ Slope of the Tangent to a Curve 
3. A curve with a corner at 
P(2, 2) is given. 
(a) Show that OP lies on the 
line y = x. 
(b) Show that PT 1 ies on the 
line x + y = 4. 
(c) Find the coordinates of R 
given its x coordinate 
P (2,2) 
yV 
T 
o 
(2 - h). 
(d) hind the coordinates of Q given its x coordinate (2 + h). 
(e) Find the slopes of the secants PR and PQ. 
(/) If a tangent exists at P, sandwich its slope value between 
two numbers. 
(g) Can the tangent slope be found as h —>0? Explain. 
4. A curve is defined as follows: 
y = x2, x > 0 
y = x2 — 1, x < 0. 
This curve has a break at 
x = 0. 
(a) Find the coordinates of R 
given its x coordinate 
(0 - h). 
(b) Find the coordinates of Q 
given its x coordinate 
(0 + h). 
(c) Find the slopes of the 
secants PQ and PR. 
(d) If a tangent at P exists, 
sandwich its slope value 
between two numbers. 
(e) Can the tangent slope be 
found as h —» 0? Explain. 
5. A curve is defined as follows: 
y = (x — l)2, x > 0 
y = (x + l)2, x < 0. 
This curve has a corner at 
P(0, 1). Repeat Question 4 
for this curve and tangent atP. 
y = 
1.2/Slope of the Tangent to a Curve 9 
1.3 Limits 
In Section 1.2 we “sandwiched” the value of the tangent between 
two expressions such as 4 + h and 4 — h, and found that their 
values could be brought as close together as we wished. You may 
wonder why we let h —> 0 instead of putting h = 0 directly. Notice 
that in obtaining the values 4 + h and 4 — h we had to divide by 
h and for this reason we had to stipulate that h + 0, since division 
by zero is not defined. 
Limits such as lim 4 + h are easy to find because to get the limit 
A-*0 
we can actually put h — 0. Here are some limits that can be obtained 
by direct substitution. 
lim x = 2 
x—*2 
lim x + 3 = 7 
jr-»4 
lim 19 = 19 
x—*2 
lim 
X—>1 
x + 1 
x — 3 
-1 
x2 - 3x + 5 1-3 + 5 3 
lim- =- = - 
x-, 2x + 5 2 + 5 7 
lim (x — 2)(x + 5) = 1X8 = 8 
z->3 
Substituting for the variable is not always possible. For example, 
lim - would be —, which is undefined. 
x-tO X 0 
To study lim - we look at the following table, where x takes on 
x—► 0 % 
positive values 0.1, 0.01, . . . and negative values —0.1, —0.01, . . . . 
X 0.1 0.01 0.001 0.0001 -0.1 -0.01 -0.001 
1 
X 
10 100 1000 10 000 -01 - 100 -1000 
The table shows two cases for x approaching 0: x—> 0+, which 
means that x approaches 0 through positive values, and x —»0~, which 
means that x approaches 0 through negative values. 
As x * 0+ the number - becomes arbitrarily large. That is, - be¬ 
comes larger than any positive number. 
As x —>0“, the number-is negative and becomes arbitrarily large 
in magnitude. 
We express this situation by writing 
lim -|— = 03(or + 03) and lim — - = — co 
z->0 x x_>0 X 
Note that: lim - cannot be expressed as a real number. 
10 1.3/Limits 
Also, observe the symbol00 indicates a process involving arbitrarily 
large numbers. The use of this 00 symbol should conform to the nature 
of such processes since if conceived as a number symbol, it does not 
share all the properties of real numbers. (Read the chapter “Beyond 
the Googol” in Mathematics and the Imagination by Kasner and 
Newman, Simon N Shuster.) 
Thus it is possible to interpret lim - = <» as a shorthand way of 
*o+ x 
saying that as x becomes arbitrarily small, though positive, - becomes 
arbitrarily large. X" 
Remember that <=° is not a name for any real number. The symbol 
<0° ’ signifies a process of becoming arbitrarily large. 
Example 1 may be contrasted with another important limit, 
lim 1 = 0 
2—►co X 
Here are some further cases of limits in which direct substitution 
leads to expressions that do not name real numbers. For instance, 
consider the function defined by 
/(*) = x - 2 ’ X ** 2‘ 
To find 
lim fix) = lim ~ , 
x->2 z->2 x L 
we do not consider the value of /(2) (which has not even been defined!). 
We consider instead the values of /(x) for x in the neighbourhood of 2, 
but different from 2. 
x2 - 4 = (x + 2)(x - 2) 
Since x / 2, we can divide by x — 2. 
v2 — A 
- = x + 2, x 9^ 2 
x - 2 
lim —-— = lim (x + 2) 
a;—* 2 X 2 x—> 2 
= 4 
The value of the limit is 4. 
If we had attempted to evaluate this limit by direct substitution 
of x = 2 in 
x - 4 
x — 2 
, we would have obtained the meaningless symbol # . 
Such an expression is called an indeterminate form. 
1.3/ Limits II 
Example 1 
Find lim 
x->0 
2x2 + 5x 
x 
Solution 
Substitution yields the expression $ which is undefined. 
But 2x + 5x 
2x2 + 5x 
x 
lim 
2x + 5x 
x 
x(2x + 5) 
2x + 5, x 
lim (2x + 5) 
£->0 
5. 
0 
Notice that as x approaches zero, x never assumes the value zero. 
Zero has been excluded from the domain of the above algebraic 
identity because division by zero is not defined. 
Example 2 
Find lim 
X—*co 
2x + 3 
5x — 7 
Solution 
As x gets larger so do 2x + 3 and 5x — 7. The value of the limit 
is not apparent. One limit involving oo that is known is 
lim - = 0, that is, - —>■ 0 as x —»®. 
x—►^o oc oc 
If x 7^ 0, we may divide numerator and denominator by x. 
2-L- 
2x + 3 ^ x 
5x — 7 
5-? 
X 
As x 
as x 
lim 
x-^co 
X 
2x + 3 } 2 
5x — 7 5 
2x + 3 _ 2 
5x — 7 “ 5 ' 
and 
and 
2 + - —> 2 
x 
x 
12 1.3/ Limits 
Example 3 
Find 
2x2 -3x+ 1 
™ 5x2+ x-7 ■ 
Solution 
We shall transform this expression so that we can use 
lim- 
'T-Xrv-, 00 
o. 
We divide both numerator and denominator by x2, x ^ 0. 
2 2-^.Llim fy- 3* + 1 = |im . * 
£-}CO 5x + x — 7 £->co 
2 
5 
5 + - 
x x 
Notice in Example 3 we divided numerator and denominator by 
the highest power of x in the denominator. 
Example If 
Determine lim — ^-— . 
Solution 
Substitution of h = 0 yields an indeterminate form However, 
(a + h)2 - a2 = o2 + 2ah + h2 - a 
= 2ah -f- h 
(a + h)2 — a2 2 ah + h2 
h ~ h 
= 2a + h, h 9^ 0 
lim ^-— = lim (2a + h) 
o n h-> o 
= 2a 
Observe that the algebraic identity used in this calculation is not 
valid for h — 0. However, the value zero is not actually taken by h 
as h approaches zero. 
1.3/ Limits 13 
Exercise 1.3 
1. State the limits of each of the following expressions as x ap¬ 
proaches 1. 
2. 
3. 
4. 
(a) (c) 
X 
X 
1 + X 
(b) (d) 
1 — X2 
X 
1 + X2 
State the value of each of the given limits. 
(a) lim x6 (e) lim (a2 + 2ah) 
z—>0 /i-*0 
(b) lim \/x, (x > 0) (/) lim (5 + 6/z) 
z-*0 
(0 lim x(x + 2) 
z—>3 
(g) 
.. x — 10 
lim 
x—>10 x + 10 
(<*) lim (x4 + x2), a £ R (h) lim cos x 
x—*a x—* 0 
By inspection, state the value of 
lim/(x) and lim/(x), where the 
x—► 1 x~* 4 
graph of / is as shown. 
Does /(x) have a limit at x = 2, 3, 
If not, explain. 
y 
4 
O 
o 
1 \ 
o 1 2 3 i 5 
4? y 
5. Determine the values of the 
(a) lim (3y — y) 
2/-»0 
(b) lim 
o 
(c) lim 
0 
1 +1 
1 - t 
2 i 
Z + Z 
2 | -j 
2 + 1 
given limits. Graph of / 
(d) lim (w2 — aw) 
0 
(<e) lim (sin r) 
r~>0 
(/) lim 2s 
s-» 0 
14 4.3/ Limits 
6. Find the value of each of the following limits. 
(a) lim | 
.... x — 16 
(c) hm -— 
x->4 X — 4 
(b) lim 
x - 9 
(J) lim 
£->0 £->—3 3C ~f" 3 
7. Determine the value of each of the given limits 
^2 r.2 
h 
(* + 1) - 1 
x 
(.) lim <2 + A> -2 
h-i 0 
.... (7 + 4)' - 7’ 
(c) 5 (7 + 4)2 - 7* 
(d) lim 
A 
h^o (3 + h) — 3 
M Hm -<2 + «>’ - 21 
:r-»0 ^ 
(/) lim (5 + *)‘ ~ 5« 
*->o n 
8. Evaluate the following limits. 
(o) 1” TJi 
(b) lim 7*; J'5 
z^co ^ 
(c) lim 
X-+CD 
2xz - 5x + 7 
3x2— 1 
(d) lim 
(e) lim 10 
Z-KO 
(/) lim 10 
3—x 
9. If x—* 2+ means x > 2 and x —» 2, then x may take on values 
such as 2.1, 2.01, 2.001, .... Also x —> 2~ means x < 2 and 
x —> 2, so that x may take on values such as 1.9, 1.99, 1.999, .... 
Find 
/ \ i • x “f~ 3 
(a) lim -- 
v ' ^2+ x - 2 
/i \ i • x ~F 3 
(b) hm_ Z-o x->2 X 2 
10. Find 
(a) lim 
z-»<r 
1 - 2~i 
1 + 2~3 
(b) lim 
z-^Q 
1 - 2- 
1 + 2- 
11. Evaluate the given limits, where a,b, c £ R. 
(a) lim + 2f - a‘ 
h-> o n 
(*) lim <* + R - 4a* 
x->a % Cl 
(c) lim 
x-^a 
(d) lim 
(2x — a)2 — a2 
x — a 
2 
x — c 
\/x — c 
1.3/Limits 15 
1.4 Limit of a Sequence 
Two profound ideas underlying the work of Archimedes, Newton 
and Leibniz required further development and clarification to give 
calculus the logical precision enjoyed by most branches of mathe¬ 
matics. The first of these basic concepts can best be illustrated by 
one of Zeno’s paradoxes. (Zeno was a Greek philosopher, 495-435 
B.C.) 
A hare and a tortoise decided to have a race. The confident hare 
gave the tortoise a 100 m handicap. Assuming that the hare runs 
10 times as fast as the tortoise we may however argue that the hare 
can never overtake the tortoise for the following reasons. 
(1) W1 len the hare runs 100 m the tortoise is 10 m ahead. 
(2) When the hare runs 10 more metres the tortoise is 1 m ahead. 
(3) When the hare runs 1 m further the tortoise is 0.1 m ahead. 
—and so on forever for a never terminating sequence of similar stages. 
In spite of all their genius for mathematics, the ancient Greeks were 
unable to see the fallacy in this argument. The sequence of numbers 
describing the distances run by the hare or the tortoise, namely, 
100, 10, 1, 0.1, 0.01, . . . 
is infinite, and has no last term. I lie hare can nonetheless overtake 
the tortoise at a definite instant of time, because, if the hare runs at 
the rate of 1 m/s, the sequence of numbers measuring the total 
time taken by the hare, namely 100, 110, 111, 111.1, 111.11, 111.111, 
. . ., is a sequence that has a limit. Therefore, when the time has 
passed this limit, the hare has passed the tortoise. 
16 1-4/Limit of a Sequence 
The paradox of Zeno became fully explained and clarified only after 
the work of three German mathematicians of the nineteenth century. 
Bernard Bolzano (1781—1848), who was professor of the philosophy 
of religion at Prague, wrote a book “Paradoxes of the Infinite” which, 
alter its posthumous publication in 1851, became recognized as a 
masterpiece of logical and mathematical thought. Ix. L. Weierstrass 
(1815-1897), one of the cleverest mathematicians of his time, was a 
high school teacher in Munster until 1856 when he became a pro¬ 
fessor at Berlin. (>eorg Cantor (1845-1918), the founder of the modern 
theory of sets, was born in St. Petersburg (Leningrad), studied in 
Germany, and became a professor at Halle in 1879. To these scholars 
we owe much of our present understanding of the concepts of sets, 
sequences and limits. 
Mathematicians employ a variety of equivalent definitions of 
an infinite sequence. Logically, the simplest is the following. 
An infinite sequence is a function defined on the natural 
numbers N. 
For example, the function 
is an infinite sequence. T 
n y/n 
0 0 
1 
2 1.41 
3 1.73 
4 2 
5 2.24 
6 2.45 
7 2.66 
8 2.82 
9 3 
-» -\/n where n £ N 
e graph of / follows. 
y/n 
Graph of/: n —> y/n 
n 
1.4/Limit of a Sequence 17 
An element of the range of the sequence is called a term of the 
sequence. 
The elements of the sequence can be arranged in order according 
to the natural order of N. Thus, in the above example, the terms of 
the sequence arranged in their natural order are 
Vh \/2, \/3, V4, ’» • • • 
Thus, when the elements of a sequence are distinct, the sequence can 
also be described as an ordered set. 
By the phrase, the kth term, tk, of a sequence f, we mean the value 
f(k) which the function associates with the natural number k. 
If the domain of / is a finite subset of N, we call / a. finite sequence. 
In this case, although not logically necessary, it is customary to take 
the domain to be the initial interval of N. For example, if / is defined 
for only ten natural numbers, we would use {1, 2, 3, ... , 10} as the 
domain and speak of a finite sequence of ten terms. 
Sequences are most frequently described by means of the defining 
equation of the corresponding function. For example, the function 
fin-^f(n) = - 
is a sequence. This sequence can also be described by listing its terms 
in order 
1111 
1 ’ 2 ’ 3 ’ 4 » • • • 
Following common usage, we shall frequently refer to such a list of 
terms as the sequence. 
Any finite sequence has a smallest and largest term. This property 
does not necessarily hold for an infinite sequence. For example, the 
sequence 2, 4, 8, 16, 32, . . . , 2*.k £ N, has no largest term. 
Consider the sequence given by n —> tn, where 
ln JQTi ) fi £ N. 
The sequence is 
0.1, 0.01, 0.001 
f • • • 
18 1-4/Limit of a Sequence 
Notice that as n gets larger, tn gets smaller. Although there is no 
number n £ N for which tn is zero, nonetheless, tn is arbitrarily close 
to zero for sufficiently large n. 
The above sequence has the property that for any number, no 
matter how small, all but a finite number of terms of the sequence are 
closer to zero than the given number. For example, let us select the 
1 -6 
small number ^ qqq qqq or ^ ^ but t^le ^rst s*x terms are closer 
to zero than 10-6. That is to say, all but the first six terms lie in the 
interval from (0 — 10-6) to (0 + 10~6). 
How manv terms are closer to zero than 10 12? 10 18? 10 "? The 
sequence tn is said to have the limit 0. f(n) 
Geometrically, if we graph 
the function f:n —> we ob¬ 
tain a set of points that lie 
closer and closer to the hori¬ 
zontal axis as n becomes larger. 
Notice that all but a finite num¬ 
ber of points he in the shaded 
strip of width e, no matterhow 
small e is. 
For the graph of the function 
f:n—> 3 + (~To)n we n°tice that 
the strip must be drawn to en¬ 
close the horizontal line x = 3. 
Thus, for any number e, no 
matter how small, all but a 
finite number of points lie in the 
strip of width 2e; that is, be¬ 
tween the lines x = 3 + e and 
x = 3 — e. 
In general we define the limit 
of a sequence as follows. 
For a sequence n —> tn, the number L is said to be the 
limit of the sequence if, no matter what (small) positive 
number e is chosen, all but a finite number of terms of the 
sequence lie in the interval between L — e and L + e. 
l.f/Limit of a Sequence 19 
An alternative description of this definition of limit is the following. 
If for every positive number e, no matter how small, we have 
IL -tn | < e 
for all but a finite number of terms tn of the sequence, then L is said 
to be the limit of the sequence. 
Check that the two forms of the definition are logically equivalent. 
Exercise 1.4 
1. State the limit of the sequences whose terms are given in each of 
the following. 
(a) 2, 1*, A li 1 +-, 
2 3 4 n 
M 1 1 1 J_ 1_ 
C 2’ 4’ 8’ 16.2n’ 
(b) 
1 1 
3’ 4’ 
1 
5’ 
(-l)n 
n + 1 
+i 
• (d) 1, 1, 1, 1, 1, 
2. For a sequence whose nth term is 7n find the following. 
(a) the first term (c) /200 
(b) the 20th term (d) tk 
3. List the first five terms of the sequences determined by the 
following functions. Variables have domain N. 
(a) /: x ->/(*) = x [d) /: k = k2 
(b) g: k —> g{k) = 2k + 1 (e) g: s g(s) = —-— 
s + 1 
(c) h: n —> h(n) = 3n — 1 (/) h:m-*h(m)=~ 
w3 
4. C onstruct the first six terms of the sequences defined by the 
following formulas. 
(a) t\ = 1, tn = 3tn — i n C N, n \ 
(■b) h = 2, 4 = 4 , ! + 3k k £ N, k y± \ 
(c) h = —3, tn = tn _ x — In n £ N, n l 
5. (a) Graph the infinite sequence f:n -n £ N 
n + 1 
(b) How many points of the graph lie outside the strip defined 
by {(x, y) | 0.999 < y < 1.001}? 
20 1.4/Limit of a Sequence 
6. (a) Graph the infinite sequence /: n —> 3 + (—|)n, N. 
(b) Describe a strip that contains all but a finite number of 
points of the graph of/. 
7. Find the first five terms of each of the following infinite se¬ 
quences. Suggest the limit of each. 
(«) /: n —> -3 + n 6 N 
(b) q: n —> 2 + n 6 iV 
n 
2n + 3 
(c) f:n—>-, n d N 
n 
(d) h: n —> —-—, n £ N 
w F 1 
/ \ "F 1 
(e) n —> -—, n 6 N 
1 — Sn 
8. The terms of a sequence are given as follows. 
tx = 8, and for n > 1, tn+1 = \tn + i 
(a) Find the first five terms of the sequence. 
(.b) Suggest a limit for the sequence. 
9. (a) Find the first five terms of the sequence defined as follows. 
tx — 2, and for n > 1, tn+1 = \/Stn + 4 
(ft) Conjecture the value of lim tn. 
(c) Show that if t = lim tn then 
t2 — St — 4 = 0 
1.5 Sums of Infinite Series 
We are now ready to describe the second idea that underlay the 
work of Archimedes, Newton, and Leibniz. This is the perception 
that the sum of an infinite series could also be described by means of 
a limit. The successive distances run by the hare are 100, 10, 1, 
tL, .... Although infinite in number, the total distance represented 
by 100 + 10+1 + iV + ... is finite. How can we determine the 
value of such a sum? Recall that a, ar, ar2, ar3, . . . , arn~x is a finite 
geometric sequence with first term a and common ratio r. The 
indicated sum a + ar + ar2 + . . . + arn ~ 1 is called a geometric 
series. The sum of the geometric series is 
a(l — rn) 
v = _A-L 
^ n . 
1 — r 
1.5/Sums of Infinite Series 21 
If the terms of a series continue indefinitely so that there is no 
last term, then the series is called an infinite series. How can we find 
the sum of an infinite geometric series such as a + ar + ar2 + . . . ? 
To answer this question we consider the partial sums 
51 = a 
52 = a + ar 
53 = a + ar + ar2 
Sn = a + ar + ar2 + . . . -fi arn ~ 1 
Notice that Si, S2, .... S„, . . . themselves form a sequence that 
is called a sequence of partial sums. 
If this infinite sequence of partial sums has a limit L, then we shall 
consider L to be the sum of the infinite geometric series. 
Example 1 
For the infinite geometric series 
1+2+i+8+T6+--- 
(a) Find and graph the first five partial sums. 
(b) Find the sum of the series. 
Solution 
(a) Si = 1 
5, = 1- 
2 
53 = 1- 
54 = 1 
S6 = 1 
3 
4 
7 
8 
15 
16 
Sn 
O 
• • 
12 3 4 5 
22 1.5/Sums of Infinite Series 
aiX — rn) 
1 - r 
(b) 
Sn = 
Id ~ (*)") 
1 - i 
lim ^zrr = 0, lim Sn = 2 
n-^co ^ n-$co 
the sum of the infinite geometric series is 2. 
We shall denote lim Sn by Sm. 
71—^CO 
Problem 
For the general geometric series 
a + ar + ar2 + ... 
find the condition on r such that Sm exists, and find Sm. 
Solution 
Recall, the sum of n terms of the geometric series is 
Sn 
_ a( 1 — rn) _ a _ ar11 
1 — r 1 — r 1 — r ' 
The behaviour of Sn for large n depends entirely on the behaviour 
of rn. 
n 
CLT 
If \r\ <1, then as n —> », rn —>► 0 and thus, ^ _ r Therefore, 
exists and its value is 
a 
1 - r ‘ 
If |r| > 1, then lim rn does not exist and does not exist. 
rc-»oo 
For r = 1, the formula for Sn fails. But the series becomes 
a + a + a + a + ... + a + ... 
Sn = na 
and Sw does not exist. 
1.5/Sums of Infinite Series 23 
por 7 = _i the series becomes 
a-a + a-a + a-a + ... • 
and 
S„ = a 
Sn = 0 
if n is odd, 
if n is even, 
does not exist. 
Example 2 
Find the sum of the infinite geometric series 
9 + 3 + l + l + i + -- - • 
Solution 
For a geometric series with |r| <1, 
a 
S' = - 
I — r 
1 
a = 9 , r - g 
27 
2 
Example S 
Express the rational number corresponding to 
0.723 
as a quotient of integers. 
Solution 
0.723 = 0.723 + 0.000 723 + 0.000 000 723 + . • ■ • 
1.5/Sums of Infinite Series 
This is an infinite geometric series, with 
a = 0.723 , r = 0.001 
V M < 1 , S.= ~— 
1 — r 
« = 0.723 
“ 1 - 0.001 
= 0.723 
0.999 
241 
~ 333 
Therefore, -ff-J and 0.723 are different numerals for the same number. 
Exercise 1.5 
1. Express each of the following periodic decimals as an infinite 
series. 
(a) 0.7 (b) 0.62 (c) 0.524 (d) 5.6 
2. Express each of the following as the sum of a number and an 
infinite geometric series. For example, 
1.723 = 1.7 + 0.023 
= 1.7 + 0.023 + 0.000 23 + . . . 
(a) 1.3 (b) 3.56 (c) 0.17 (d) 1.236 
3. Find the sum of each of the following infinite geometric series. 
, v t 1,1 1 , (a) 1 - - + i - - + . . . 
(b) 6 + 2 + - + -+ ... 
(c) 9 — 6 + 4 — - + ... 
... . 120 . 360 . 
(d) 40 + — + — + 
(e) \/6 + 2 + 2 /|/^ + 
4. Find the first three terms of the infinite geometric series given the 
data in each of the following. 
(a) a = 3, Sm = 3| (5) r = §, S„ = 36 
1.5/Sums of Infinite Series 25 
5. Express each of the following as an infinite geometric series. Find 
the sum, and express it as a quotient of integers. 
(a) O.i (c) 0.12 (e) 0.83 
(ib) 0.86 (d) 1.45 (/) 0.0081 
6. A frog on a log, 1 unit long, jumps half way to the far end, then 
again half the remaining distance, again, and again, and so on; 
his jumps forming the sequence J, j, j, . . . . 
(a) Find the distance jumped in the wth jump. 
(b) Find the limit of the jump length as n —»®. 
(c) Find the total distance jumped by the first n jumps. 
(d) Find the limit of the total distance jumped as n —» ®. 
7. Repeat Question 6 for Mr. B. Frog who jumps 3 of the way to 
the end of the log with each jump. 
8. Express each term of the infinite sequence 
VS, V-3 x/3, V3 V3 \/3, ^3 V.3 V3^,. .. 
as a power of 3. Show that the sequence has a limit, and find the 
limit. 
9. (a) Determine 
S' = Drn 
1 
+ 
1 
+ 
1 
n + 1 (ft + l)2 (ft + 1) 
+ . . . , where n > 0. 
(b) Determine if exists if n =■ 0, —1, or —2. 
10. Given the series 1 + 1+1+.. . + 1. 
(a) Can this series be considered as geometric? Explain. 
(b) Can the formula for the sum of a geometric series be applied 
to find the sum of n terms of the series? Explain. 
11. Repeat Question 10 for the series1 — 1 + 1 — 1+.... 
12. (a) Find the sum of the series 1 + a + + + ... + a\ a + 1. 
(b) Show that (1 - o)(l + a + a2 + . . . + a?) = 1 — a8. 
13. (a) Find the sum of the series 
1 + cl + 0} + . . . + au, a + 1. 
(b) Simplify (1 - a)(l + a + a2 + . . . + a24). 
26 1.5/Sums of Infinite Series 
14. (a) Find the sum of the series 
x® + ax8 + a2x7 + . . . + a9, x + a. 
(b) Show that (x — a)(x® + ax8 + . . . + a9) = x10 — a10. 
15. (a) Find the sum of the series 
x"-1 + ax'1-2 + a2xn~3 + . . . + a”"1, x + a. 
(b) Simplify (x — aXx"-1 + ax"~2 + a2xn~3 + . . . + a”-1). 
16. Factor the following expressions. 
(a) a6 — 1 (b) x11 — a11 (c) a8 — b8 (d) an — bn 
17. The sum of the first two terms of an infinite geometric series is 5. 
Each term is 3 times the sum of all the terms that follow it. Find 
the sum of the series. 
18. The midpoints of the sides of a triangle are joined to form a tri¬ 
angle. The process is continued for each new triangle formed. 
Show that the area of all new triangles so formed is one third 
the area of the original triangle. 
1.6 Special Series 
Two basic ideas, limits and series, have played an important role 
in our development of the calculus. We shall introduce a useful 
notation and several important sums of series in this section. 
For example, consider the series 2 + 4 + 6 + 8 + 10 + 12 + 14, 
which may be rewritten as 2(1) +2(2) +2(3) +2(4) +2(5) + 
2(6) + 2(7). The &th term of the series is 2k, since every term can be 
obtained by putting k = 1, 2, 3, . . ., 7 successively. This series has 
a sum that may be described as “the sum of terms of the form 2k for 
k = 1, 2, 3, . . ., 7” or simply as “the sum of 2k for k from 1 to 7.” 
We abbreviate these expressions by using the notation 
7 
£ 2k = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + 2(6) + 2(7) 
4=1 
where the Greek capital letter S(sigma) corresponds to the word 
“sum.” The letter k is called the index of summation. 
1.6/Special Series 27 
Example 1 
Write the following sums explicitly. 
(a) £ k2 (b) £ 2(30 
*=i i=i 
Solution 
5 
(а) Z k2 = l2 + 22 + 32 + 42 + 52 
A=1 
6 
(б) Z 2(30 = 2(3) + 2(32) + 2(33) + 2(34) + 2(35) + 2(36) 
2 = 1 
Notice that in Example 1(6) we used i as the index of summation. 
Example 2 
7 
State explicitly the series indicated by Z (2k - 1) 
A-=3 
Solution 
7 
£ (26 — 1) = (2 X 3 — 1) + (2 X 4 — 1) + (2 X 5 — 1) + 
4=3 (2 X 6 - 1) + (2 X 7 - 1) 
= 5 + 7 + 9 + 11 + 13 
= 45 
In the exercise that follows you will find the sums of two important 
series 
n 
(1) ^ — 1+2 + 3+ 4+ ... n 
i= 1 
n 
(2) Z *2 = l2 + 22 + 32 + . . . + 7z2 
2=1 
Exercise 1.6 
1. \\ rite each of the following sums explicitly. 
(a) Z * (c) Z 63 (e) Z f(i) 
2=1 A=1 
(b) Z3i (d) Z(-l)^2 
*=i 
28 
(/) Zt-W2-*) 
2=1 
1.6/Special Ser ies 
2. Express each of the following series using the sigma notation. 
(■a) 3 + 6 + 9 + 12 + 15 + 18 
(b) 1+2 + 4 + 8 + 16 
{c) 1 + l + i + I + Ti + 2 
(d) 2 + 6 + 18 + 54 + 162 
(e) 2 - 6 + IS - 54 + 162 
3. Show that the following pairs of expressions name the same 
k + 3 
*=o \k + 5 
3 'i + 5 
series. 
(a) 
6 
E i= 3 (rh) and 
(b) 
II 
( j \ and 
V + -4/ 
(c) 
n 
E <2t + l and 
i=0 \^ 
n+1 
4. Find the sum of the series X ( —1)* if n is odd. If n is even. 
A-= 1 
5. Black dots may be used to represent sums of the form yX 
• • • 
• • • 
• • • • 
• • • • 
• • • 
• • • • • 
• • • • 
• • • • • 
• • • • • 
1+2 
2X3 
2 
1+2 + 3 
3X4 
2 
(a) Represent each of 1, 1+2, 1+2+3, 1 +2 + 1 
sigma notation. 
5 6 
(b) Use the pattern suggested to find X i and X z- 
!=1 i= 1 
(c) Prove X i — n(n + 1) 
i=i 
1.6/Special Series 29 
6. Dots are used to represent sums of the form ^ (2i — 1) 
1 
• • f t • 
• • • 
• • • 
• • • 9 
• €> • • 
• • • • 
• • • • 
1+3 1+3+5 1+3+5+7 
= 22 = 32 = 42 
(a) Represent each of 1, 1+3, 1+3+5, 1 + 3 + 5 + 7 in 
sigma notation 
(b) Use the pattern suggested to hnd (2f — 1) and (2f — 1) 
i= 1 i=l 
(c) Prove ^2 (2i — 1) = n2. 
i= i 
7. Dots are used to represent sums of the form 
I2 + 22 + 32 + 42 
l2 = 1 
22 = 1 + 3 
• 
4 X 1 
32 = 1 + 3 + 5 • 
42 = 1+3 + 5 + 7 • • • 
• • • 3X3 
4 
• • « • • 
2X5 
Et2 = 4Xl+JX3 + 2X5+ X 7 
1 
1 X 7 
(a) Show that the rectangular array 
4 
of dots represents 3 ^'2- ::: t 
(b) Draw a rectangular array of dots 
CN 
5 
to represent 3 22 i2. 
i= 1 
7 +2 
4 5 
(c) Evaluate ^ i2 and "22 i2. 
»— 1 i— 1 
30 1.6/Special Series 
(d) Draw a rectangular array representing 
n 
3 £ f2 and show that the number of dots 
1=1 
in the array is 
(1 + 2 + 3 + . . . + n)[{2n — 1) + 2]. 
n 
(e) Show that ^ ^ ~ ~kn (n T l)(2n + 1). 
i=i 
8. I3 = 1 
2-3 = 2 + 4 + 2 
33 = 3+ 6 + 9 + 6 + 3 
43 = 4 + 8 + 12 + 16 + 12 + 8 + 4 
These sums can be arranged as follows. 
4 
£ i3 = l3 + 23 + 33 + 43 
jj 2 3 4 
2 4 6 8 
3 6 9 12 
4 8 12 16 
1 
(a) Make an array to show £ i3. 
i=i 
ib) Express the sum of each row of the array as 
6(1 + 2 + 3 + 4 + 5),k£ N. 
(c) Show that 
± = 1 ± i + 2 £ i + 3 ± i + 4 £ i + 5 £ i, 
t-1 1 1 1 1 1 
5 5 
where ^ i means ^ i 
i t= i 
5 
= (1 + 2+ 3+ 4 +5) ^ i 
(d) Draw an array of numbers representing £ iz and 
i= 1 
show that £ i3 = ( £ 
i=i 
n(n + 1) 
1.6/Special Series 31 
9. Show that 
5 
(a) 22 a = 5a (b) Eo = na 
i= 1 2=1 
10. Show that the following equations are true. 
(a) £ (9oi) =9(Z *) 
i—1 M=1 / 
4 4 4 
(b) 22 (°* + b%) = 22 ai + Z 
i= 1 2=1 2=1 
12 
(c) 22 («*+i - = «i3 - «i 
2=1 
n 
(d) 22 (fli+l ~ fl») = «n+l - <2l 
2=1 
(e) 2 (r + l)2 - i2 = (» + l)2 - l2 
= «2 + 2« 2= 1 
(/) Z (» + l)2 - 22 = Z (22 + 1) 
(g) Z (22 + 1) = 2 Z* +n 
2=1 ' 2= 1 / 
11. Use parts (e), (/), and (g) of Question 10 to show that 
n 
2 Z 2 = w2 + n 
n{n + 1) 
i=i 
or 1 + n = 
12. (a) Show that Z \(i + l)3 — t3l = (n + l)3 — 1 
i=i 
(b) Use the result of (a) to show that 
n3 + 3«2 + 3w = Z (3+ + 32 + 1) = 3 Z + + 3 22 i + 
*=1 i= 1 i—l 
(c) Use the result of (b) and Question 11 to show that 
22 i* = ”(n + 1)(2» + 1) 
n 
32 1.6/Special Series 
n 
13. Use ^2 [H + l)4 — i4] and the results of the previous questions 
i— 1 
n 
to show that X i3 
»=i 
n(n + 1) 
~2~ 
14. Evaluate the following. 
(a) X (3J + ?) 
t = 1 
40 
00 X (_ 2i’ +5) 
(c) E (4^- + 7) 
2 = 1 
ld) X {(2i + !) 
1.7 Calculation of Areas 
Returning to the problem of finding areas of Section 1.1, we will 
show in greater detail how sequences, series and limits can be used 
to calculate the numerical values of given areas. Although other 
schemes are useful in certain instances, we shall base our calcula¬ 
tions upon the method of rectangles. The area considered is between 
the graph of a function and the x axis, and the x interval is divided 
into n equal parts, each of which is the base of a rectangle that 
extends upward to the curve. The rectangle tops may lie below 
the curve or extend above it. We obtain expressions for the desired 
area by summing the series of rectangle areas, and taking limits 
as n becomes large. 
12 
00 X - w*'+ 2) 
i=i 
50 
2 = 1 
(/) X + !)0 -2) 
1 _ _J__N 
i 1 + i/ 
10° / 1 ] 
(«> X 
2 = 1 
2=1 
0) X V2T+1 - V2i- 
1.7/Calculation of Areas 33 
Example 1 
(a) Find the sum of areas of rec¬ 
tangles of width — inscribed 
n 
in the triangle below y = x, 
and above the x axis, where 
0 < x < 1. 
(b) Calculate the limit of the sum 
in (a) as n approaches infinity. 
Explain your results. 
Solution 
y 
(a) There are n — 1 rectangles with heights 
1 
L 2 
n' n 
n — 1 
n 
from left to right. Since each width is —, the area sum is 
n 
An 
n 
- +- + 
.n n 
+ 
n 
n 
(b) lim An 
n-*co 
= 1 im 
= -2(1 + 2 + 
nz 
i n—1 
i 
1 1 r 
= ~2 • 
1- 
n 
2 
1 - - 
n 
= \ lim l 1 
^ n-> co 
l 
~ 2 
+ n — 1) 
1) 
As n increases, the number of rectangles increases, and the width 
of each rectangle decreases. The difference between the area we 
pick and the rectangle areas is a “saw-tooth” region whose “teeth” 
become smaller as n becomes larger.Thus, the limit of the sum of 
the rectangle areas as n —> °° will be the area of the triangle, and 
this limit is 
34 1.7/Calculation of Areas 
To find areas by this method, we must be able to sum the series of 
rectangle areas. For each curve this poses a special problem. Among 
the ancients, Archimedes showed the greatest skill in overcoming 
such difficulties. 
If the curve defining the upper boundary is the parabola y = x2, 
then we would need to find sums of squares such as 
F + 22 + . . . + n2 = ^2 i2, as in the next example. 
i=i 
y 
Example 2 
(a) Find the area sum of rec¬ 
tangles of width - drawn as 
n 
shown on the parabola y = x2 
where 0 < x < 1. 
(b) Calculate the limit of the sum 
in (a) as n approaches infinity. 
Explain. 
Solution 
O 2 2 3 4_ 
n n n n 
n-3 n-1 
n 
(a) Each rectangle height is the y coordinate of a point on the 
parabola y = x2 at the upper right hand rectangle corner. The 
x coordinates of these points are —, —, ...,—, so their y coordi- 
n n n 
nates are f — 
\n n n 
— I — ) . Since each rectangle has width —, 
the sum of rectangle areas is 
2 /^\ 2 
An = - 
n l) + \~J + • + 
n 
n 
= (I2 + 22 + 32 + . . . + n2) 
n2 
1 n(n + 1) (2 n + 1) 
n3 6 
2 n3 + 3 n2 + n 
(See page 33) 
6 n3 
= -+--+ —- 
3 2 n 6 n2 
1.7/Calculation of Areas 35 
m I™ A- = (l+h+6^) 
1 
“ 3 
Since the portions of rectangle areas above the curve approach 
zero as w —»• », the area under the parabola is equal to the limit of 
rectangle sums, that is, y. 
For more general curves the problem of finding sums of the series 
can be very difficult. Indeed, these difficulties blocked the advance 
of mathematics for many centuries after Archimedes. However, for 
practical purposes it is often enough to know areas with reasonable 
accuracy, and we shall therefore look for a simple method of approxi¬ 
mation. 
Rather than using rectangles 
that lie below the curve, or have 
tops above the curve, we shall gain 
in accuracy if we take the average 
of these two extremes. Let us join 
the two points of the curve, as 
determined by the width of the 
rectangle, by a straight line. The 
quadrilateral thus formed with two 
parallel sides is a trapezoid. In 
the diagram, the width of the trapezoid is x2 — Xi, and the lengths 
of the two parallel sides are /(x 1) and /(x2). Consequently the area 
of the trapezoid is ^-(x2 — Xi)(/(xi) +/(x2) ) 
Consider the area beneath the curve y =/(x), where a < x < b. 
Divide the x interval into n equal parts, each of width h = — 
n 
y 
36 1 -7/Calculation of Areas 
Construct the trapezoids with upper sides joining successive points 
(xk, /(at*) ) on the curve. Then the area under the curve is given 
approximately by the sum of trapezoid areas. Since values of / at 
xn x2- ' ’ •> x«-i each appear twice in trapezoid areas, we obtain the 
following result. 
Trapezoid Rule 
= 2 U(a) “h 2/(xi) + 2/(x2) + . . +2/(xn_0 +f(b)] 
Here h —-, and a = x0, b = xn are end points of the interval. 
n 
Observe that we have assumed that /(x) has positive values, so 
its graph lies above the x axis. 
Example 3 
Calculate the area under 
the curve y = x3, 0 < x < 1, 
using the Trapezoid Rule 
with n = 10. 
Solution 
Here h = 0.1 and values 
of / are given in the table. 
X fix) = X3 
0 0 
0.1 0.001 
0.2 0.008 
0.3 0.027 
0.4 0.064 
0.5 0.125 
0.6 0.216 
0.7 0.343 
0.8 0.512 
0.9 0.729 
1.0 1.000 
1.0 
0.5 
O 
Hio = °^[0 + 2(0.001 + 0.008 + 
= ^[2(2.025) + 1.000] 
0.5 
+ 0.729) + 1.000] 
= 0.2525 
The exact area is j, so in this instance the error is about 1%. 
For greater accuracy, a larger value of n could be employed. 
1.7/Calculation of Areas 37 
Exercise 1.7 
1. Given an area with curved boundaries, describe how it can be 
divided into pieces each of which is the shape of an area be¬ 
neath a curve as discussed in this section. 
2. If An is calculated by the Trapezoid Rule, what is the value of 
lim An? 
n—*co 
3. (a) For the area beneath y = x, where 0 < x < 1, find the area 
sum for rectangles of width — with tops just above the line 
y = x. n 
(b) Compare this sum with that of Example 1. For what values 
of n do the two sums agree to 3 decimal places? 
4. (a) For the parabola y = x2, 0 < x < 1, find the sum of areas 
of rectangles of width — inscribed beneath the curve. 
n 
(b) Compare with Example 2, and state the difference in the 
two estimates. 
(c) Explain the value of this difference geometrically. 
5. (a) For the parabola y = x2, where 0 < x < a, find the sum of 
areas of rectangles of width - inscribed beneath the curve. 
n 
(b) Letting n approach infinity, deduce the exact value of this 
area. 
6. Use the method of Example 2 to find the area of the region 
defined by the boundaries in each of the following. 
(a) x = 0, y = 0, x = 2, y = x2 + 1 
(b) x = 0, y = 0, x = 2, y = x2 + 3 
(c) x = 0, y = 0, x = 2, y = 2x2 + 5 
7. (a) Using rectangles of width - below the curve y = x3 where 
n 
0 < x < 1, find An for this curve. 
(b) Calculate lim An. (See Question 8, Exercise 1.6) 
n—>oo 
8. Using the Trapezoid Rule with h =\, estimate the area under 
the curve y = 2X for 0 < x < 3. 
’8 1.7/Calculation of Areas 
9. W ith h — , apply the Trapezoid Rule to estimate the area 
below the sine curve y = sin x, where 0 < x < —. 
2 
[ Hint: — rad = 18°. 
10. (a) Show that 
^pAY = (n + l)2 
k=i\n/ 4 n 
(b) Use the result of (a) to evaluate a rectangle area sum for 
the curve y x3, w here 0 < x < 1. 
1.8 Properties of Limits 
Since calculus is essentially based on the concept of limit, we give 
here a more precise definition of the limit of a function. To say that 
a variable x approaches a number a is to consider values of x that are 
arbitrarily close to a. For a given function/, if, as x —> a, the values 
/(x) —» L, then we define lim f(x) = L. 
x—*a 
More precisely if we are given any (small) number e > 0, we must 
have |/(x) — L | < e provided that x is sufficiently close to a; that is, 
we must be able to find a positive number 5 depending on e such 
that fix) differs from L by less than e whenever x differs from a by 
less than <5. 
Symbolically, lim /(x) = L, provided that 
x—*a 
| f(x) — L | < e whenever | x — a | < 8. 
In mathematics any definition can be justified only by its usefulness. 
The precise concept of limit has proved itself to be one of the most 
fruitful definitions in modern mathematics. Much higher mathematical 
analysis of the present day rests largely on this one definition. In fact, 
differential and integral calculus, with its many applications, is only 
the beginning of its range of utility. 
1.8/Properties of Limits 39 
To make limits more useful, we must be able to conduct routine 
calculations with them. In this section, therefore, the ordinary rules 
for the use of limits are presented. There are three main rules that 
can be stated as follows, provided it is assumed that all limits referred 
to exist. 
(1) Sum rule for limits: 
lim [f(x) + g(tf)] = lim f(pc) + lim g(x) 
x->a x->a x-^a 
The limit of the sum is equal to the sum of the limits. 
(2) Product rule for limits: 
lim [/(x)g(x)] = lim /(x) lim g(x) 
x-^a x->a x-^a 
(3) 
The limit of the product is equal to the product of the limits. 
Quotient rule for limits: 
lim /(*) 
^- 
g(x) lim g(x) ’ 
if lim g(x) 0 
x->a 
x->a 
The limit of the quotient is the quotient of the limits, provided 
the limit in the denominator is not zero. 
One limit that we use frequently is the limit of a constant function. 
lim k = k. 
x->a 
We shall give a proof only for the limit rule for sums. Recall that 
\A +B\ < \A\ + \B\. 
If lim f(pc) = L and lim g(x) = M 
x->a x-+a 
then lim [f(x) + g(x)] = L + M. 
x^a 
Proof: We must show that for any given e > 0, then 
I/(*) + g(x) — (L + M) | < e 
for all x sufficiently close to (but not necessarily equal to) a 
40 1.8/Properties of Limits 
Since lim/(x) = L, and limg(x) = M, we know that for x sufh- 
x-^ax->a 
ciently close to a, we will have 
|/(x) - L\ < |, and |g(x) - M\ < |. 
To achieve this, we must use the <5 that corresponds to the small 
number 
Lj 
For such values of x, we therefore have 
I/M + *M -L-M\< |/M - L\ + | g(x) - M\ <1 + 1 = e. 
Thus, the theorem is proved. 
There are three other useful limit rules, which we shall state without 
proof. 
(1) (Inequality rule) If /(x) < g(x), then if the limits exist 
lim/(x) < limg(x). 
i->a x->a 
(2) (Sandwich rule) If /(x) < g(x) < h(x), then if the limits exist 
lim/(x) < lim g(x) < lim h(x). 
x->a x->a x-^a 
(3) (Squeeze rule) If /(x) < g(x) < h(x) and 
lim/(x) = lim h(x) = A, 
2:->a x->a 
then limg(x) = A. 
x->a 
Thus limits have many properties that conform to intuition, or 
“common sense.” When working with limits, one should be careful 
of a possible zero value in a denominator. Limit processes involve 
number sequences, and numbers are fixed, or “stationary.” How¬ 
ever, it may be useful to picture a limit process, perhaps as if there 
were an arrow hitting a bull’s-eye or a train disappearing into the 
distance. 
1.8/Properties of Limits 41 
Exercise 1.8 
A 1. State the values of the following limits, and also the limit rules 
needed to evaluate them. 
(a) lim x2 
x->a 
(b) lim x3 
x ->a 
(d) lim (3x — 2) 
x->a 
(e) lim (5x2 — 3x + 7) 
x->a 
(c) lim (1 + x) 
x-*a 
(/) Hm 
x->a 
1 — X 
1 + X2 
2. 
B 3. 
State the values of the given limits, where a, b, c 6 R. 
(a) lim 2X (d) lim x2 tan x, |&| | < ~ 
z->0 2 
(P) lim cos x (e) lim (x — x2)3* 
z-»0 
(c) lim x sin x /rt 10* 
7T 
x~*2 (f) I+ 
(a) Prove — 1 < sin x < 1 for x £ R. 
(b) Use -- < 
x ~ 
sin x 
x ~ 
-for 
X 
x > 0 and the “squeeze” rule to find 
lim 
X-^co 
sin x 
x 
4. Explain why lim 
z-»0 
COS X 
X 
does not exist. 
5. Find lim x sin - . 
x->0 OC 
Review Exercise 1.9 
1. Find the area of the two sets of rectangles that approximate the 
area of the region bounded by the parabola y = 2x2, the lines 
x = 0, y = 0, and x = 3 for the following cases. 
(a) The width of each rectangle is . 
(b) The width of each rectangle is 
(c) The width of each rectangle is y^-. 
2. Repeat Question 1 for the cubic parabola y = x3. 
42 
1.9/Review Exercise 
3. I se the method of Archimedes to find the slope of the tangent 
to y = 3x2 at (-1,3). 
4. Determine the limits of each of the indicated sequences as n—> a>, 
where n £ N. 
(a) f: n ~ 
V n 
(b) g:n->g(n) = 3~n 
(c) h: n —> h(n) = 
n + 5 
(d) k:n - 
(e) l:n- 
(/) m:n 
k(n) = 
/(n) = 
n 
n + 3 
n 
3» - 5 
2n 
m(») = ^-5 
Express as a rational number in lowest terms the sum of the 
infinite geometric series indicated in each of the following. 
(®) 1 + i + ih + • • • (c)3 — 1 + f—£ + ... 
(b) 1 — y + 4~9" — ... (d) 0.564 
6. In each of the following an infinite sequence of shaded areas is 
shown. Find the rule for forming the areas, write the sequence 
and find the sum. 
(a) _ 
1 
1.9/Review Exercise 43 
7. Evaluate the following limits. 
(a) lim (2x2 + 5x — 3) 
x->0 
oo 5 
(c) !SJ (x - 2) (x - 3) 
(^) lim 
8 
a-4—1 (3 + a)V3 - a 
2x + h 
h^o x2(x + h) 
(e) lim 772 
(/) lim 
X^co 
(g) lim 
(h) lim 
2x + 5 
x — 4 
x2 4- x — 8 
2x2 + 7x + 5 
5x2 + 11 
x3 
(i) lim 
3\n 
(b) lim [(x + 2)(x — 1) + 5] (j) lim 
X-*l 
1+ (I 
i - (ir 
(0.7)” 
(.k) lim 
1 + (0.3)” 
sin x + COS X 
tan x 
(/) lim e x, e > 1, x > 0 
£->0 
, ... x2 + 2x 
(m) lim --- 
x—> 0 OC 
in) lim 
rr-*0 
(2 + x)2 - 4 
x 
z . *. x — 7x 
(0) b37+2; 
2* 
(p) lim ri 
8. Evaluate each of the following limits. 
(a) lim 
x2 — 1 
x 4“ 1 
(b) lim (m2 — 3m) 
(c) lim 
a—* 
(d) lim 
a2 — 3a 
a—>0 0L 
3y - 7y + 7 
23,2 + 53/ + 1 
, . .. h2 - 2h 
W 
1 _ 1 
(/) limL? 
V—-3 J O 
(£> lim 
V* + 3 - 2 
x — 1 
(/z) lim 
V2 
a-»o h 
y/ a + h — 
(/) lim 
x + 1 
x — 1 
a > 0 
9. Evaluate lim (^V~+ h]-/M 
h-> 0 \ h 
(a) f(x) = x2 
for each of the following. 
(c) f(x) = x3 
(b) f{x) = 2x + 5 (d) f(x) = 
x 
44 1.9/Review Exercise 
10. Evaluate 1 im 
A—>0 
\/ a + h 
h\/ir 
(Hint: Rationalize the numerator.) 
11. By rationalizing the numerator evaluate the following. 
(a) lim 
X —►() 
V* + 2 
X 
- V2- 
(c) lim ( 
£—►+00 
(6) lim 
x—>0 
to
 
1 z
 
— X 
(d) lim 
£—► + 00 X 
x 
12. Find the sum of the areas of two sets of rectangles whose areas 
sandwich the area of the region bounded by y = x2 — 1 and the 
x axis for each of the following cases. 
The width of each rectangle is 
Write a computer program to output the sum of the areas 
of each set of rectangles where the interval is 0.01 units. 
Repeat Question 12 for each of the following regions. 
(a) the circle x2 + y2 = 100, x = 0 and y = 0. 
(b) the ellipse x2 + 4y2 = 100, x = 0 and y — 0. 
14. Write the first four terms of each of the following series. 
(a) Z 1 - k! 
Ti 1 + k2 
(b) 
n 
Z- zi 3l 
n 
(Q Z 
k=l 
1 - 2~k 
1 + 3k 
15. In the Salumbrian army every corporal commands 4 privates, 
every sergeant 4 corporals, and so on, so that every officer has 4 
immediate subordinates. The pay of every officer is twice that of 
his immediate subordinates, the daily pay of a private being one 
pestavo. There are n levels in the chain of command from private 
to general inclusive. 
(a) How many men does a general command? 
(.b) What is the proportion of privates in the Salumbrian army? 
(c) Find the total daily wage bill (in pestavos) of a general’s 
command. 
(d) What proportion in (c) is paid to the privates? 
(,e) As Salumbria rearms, n —> co . State the limits of your answers 
in (b) and (d). 
1.9/Review Exercise 45 
Slopes and Rates of Change 
Calculus is the mathematics of motion, and of change. As there is 
nothing in this world more certain than change, it should not be 
surprising that calculus has an immense variety of applications. In 
our study we shall begin with the simplest examples to illustrate the 
basic concept of differential calculus, namely the idea of a rate of 
change. 
2.1 Slope of a Linear Function 
The graph of a linear function is a straight line. Thus, the linear 
function /: x —> mx + b has as graph the straight line with equation 
y — mx -f- b. 
y 
y~m\b 
|a.v 
T Ax I 
Let (xi, yx) and (x2, y2) be any two points of the straight line. 
Recall that the slope of the segment PXP2 is defined as—, where 
Ax 
Ax = x2 — xi is the difference of the x coordinates, and 
Ay = y-i — Vi the difference of the y coordinates. 
The symbol Ax is read delta-ex”, and the reader is cautioned that 
it does not signify a product of two quantities A and x, but rather 
stands for the ‘difference of two x’s.” Likewise Ay represents the 
“difference of two y’s.” 
46 
2.1 /Slope of a Linear Function 
Example 1 
For values of x and y that satisfy the equation of the line y = 3x — 2 
complete the following table. 
with the change of x, along the length of the segment. In general, 
A y . 
— is the rate of change of y with respect to x. 
/\ -•y 
In Example 1, the value of the quotient ~ is the same for several 
Ax 
pairs of points. It would be the same for any other pairs of points 
of this straight line. The numerical value of ~ for any segment of 
the straight line is called the slope of the line. To show this we sub¬ 
tract the two equations satisfied by (xi, yf) and (x2, y2). 
V y2 = mx2 + h 
and yi = mx i + b 
y2 — yi = m(x2 — Xi) 
Ay — mAx 
Ay 
Ax 
= m 
2.1 /Slope of a Linear Function 47 
The slope of the straight line y = mx + b is m. We have shown 
that every segment (xi, yi) (x2, y2) of the straight line has the same 
slope m. 
Ay . 
Ax 
Since the value of is m, we also see that the rate of change 
of y with respect to x along any straight line is its slope. 
When the slope m of the straight line is positive, the line slopes 
upward to the right. The rate of change is positive, since it is 
equal to the slope m. Thus if the x coordinate of a point (x, y) on the 
line increases, so does the y coordinate. 
Similarly,for the line y = — ^x + 2, and any two points on it, the 
Ay . 
rate of change — is equal to the slope — Thus, as x increases, y 
decreases at ^ the increase of x. 
If the slope of a straight line is negative, the line slopes downward 
\/\] 
to the right. The ratio ~ is negative since the slope m is negative. 
When the x coordinate of a point on the line increases, the y coordi¬ 
nate decreases. 
To summarize, we see that for a straight line y = mx + b, the 
following is true. 
When the slope m is positive, y increases as x increases. 
When the slope m is zero, the line is horizontal and y is 
constant. 
When the slope m is negative, y decreases as x increases. 
Exercise 2.1 
State the slopes of the given lines. 
(a) y = 3x — 2 (c) y — 7x 
(b) y = — 5x + 4 (d) 2x + y = 5 
State the rate of change of y compared to x along each of the 
given lines. 
(a) y = — 2x + 3 (c) y = \x + 1 
(b) y = \x - | (d) x + 5y = 9 
2.1 /Slope of a Linear Function 
3. For a certain linear function /: x —> y = mx + b, — = 2 
’Ax 
(a) If x increases by 5, how does y change? 
(&) If x decreases by 3, how does y change? 
(c) If x increases by — 7, how does y change? 
(d) If y increases by 12, how does x change? 
4. For a given linear function /: x —> y = mx -j- b, — = —3. 
Ax 
(a) If x increases by 2, how does y change? 
(b) If x decreases by 4, how does y change? 
(c) If x increases by —5, how does y change? 
(d) If y increases by 6, how does x change? 
5. Find the slope of the line passing through each given pair of 
points. 
(a) (0, 1), (1,3) (c) (2, -7), (5,4) 
(b) (-2, 0), (1, 6) (d) (6, 2), (8, -5) 
6. Find the equation of the line through each of the given points, 
and having the given rate of change of y with respect to x. 
(a) (1,1); -3 (c) (-2,2); 5 
(b) (2,4); 1 (d) (-3, -2); -1 
7. Given the point P(2, —3) on the straight lineL: 4x + y — 5 = 0. 
(a) Find the y coordinate of the point Q of the line with x co¬ 
ordinate 2 + h. 
(b) Calculate Ax and Ay for the segment PQ of the straight line. 
(e) Find the slope of the line. Does the slope depend on A? 
8. Given the point P{xi, yf) on the straight line ax + by -f c = 0, 
where a, b, c £ R, b ^ 0. 
(a) Find the y coordinate of the point Q of the line with x co¬ 
ordinate Xi + h. 
(b) Calculate Ax and Ay for the line segment PQ. 
(c) Find the slope of the line. Does the slope depend on A? on xi? 
On yi? 
9. Let /: x —>/(x), x 6 R, be a linear function. Show that if 
fix -FA) — f{x) 
x, A £ R, A 5^ 0, then ^ is equal to the slope of the 
graph of /. 
2.1 /Slope of a Linear Function 49 
2.2 Slope of the Tangent Line 
Recall that given a point P on a curve and a second, variable 
point Q on the curve, then the tangent to the curve at P is related 
to the secant PQ. 
Let Q take a sequence of positions 
Qu Q2, Qz on the curve, that approach 
closer and closer to P. The secant PQ 
then takes a sequence of positions that 
also approach an ultimate, or limit, posi¬ 
tion. The tangent line to the curve at P is 
defined as the limit of a sequence of secant 
lines PQ, as Q approaches P along the 
curve. 
The slope of the tangent line is there¬ 
fore the limit of the slope of the secant. 
Let P be a given point on a curve C, and let the tangent line L 
to the curve C at P be drawn. The tangent line L has several interest¬ 
ing properties. 
Usually, but not always, L meets C at 
the point P only, whereas the secants 
PQ meet C at P and Q. 
The tangent line L may be described 
as that straight line through P that most 
closely follows the curve near P. That is, 
L is a linear (line) approximation to the 
curve C near P. 
Usually, but not always, the tangent 
at P remains on one side of the curve 
and does not cross the curve at P. The 
tangent is the only straight line through 
P that can have this property. 
C 
I he tangent line to a curve at a point P is the limit of a 
sequence of secant lines PQ, as Q approaches P along the 
curve. 
The slope of the tangent line is the limit of the slope of the 
secant. 
50 2-2/Slope of the Tangent Line 
Example 1 
Find the slope of the tangent line to the parabola y = x2 at the 
point (2, 4). 
Solution 
Let P(xi, yf) be (2, 4) and let Q(x2, y2) 
be the point on the curve with x coordi¬ 
nate x2 = 2 + h, where h is a non-zero 
real number. Then the y coordinate of 
Q is 
y% = x22 
- (2 + hy 
= 4 + 4 h + h2 
Ax = x2 — xi 
= 2 + h - 2 
= h 
Ay = y2 — y i 
= (2 + ny - 22 
= 4 + 4h + W - 4 
= 4h + h2 
. Ay 
The slope of the secant PQ is equal to — . 
Ay _ 4 h + h2 
Ax ~ h 
= 4 + h 
Now let Q approach P. Then h becomes indefinitely small, i.e. 
h -> 0. 
lim (4 + h) 
h-> o 
4 
The limit of the secant slopes is 4. Therefore, the slope of the tangent 
is 4. 
lim Al = 
A*-»o Ax 
lim A _ 
Ax->o Ax 
y 
2.2/Slope of the Tangent Line 51 
Example 2 
Find the slope of the tangent to the parabola y = 2x — x2 at 
(-1, -3). 
Solution 
Let P(xi,yi) be ( — 1, —3) and let Q(x2, yf) be the point on 
y = 2x — x2 with x coordinate x2 = — 1 h, where h ^ R, h ^ 0. 
y2 = 2x2 — x22 
= 2( —1 +h) - (-1 + h)2 
= -2 + 2h - (1 - 2h + h2) 
= -3 + 4h - h2 
yi = —3 
Aj = -3 + 4ft — h2 — (-3) 
= ih — h2 
Ax = — 1 + ^+1 
= h 
Ay _ 4h — /?2 
Ax ~ h 
= 4 - h 
Thus the slope of the secant PQ is 4 — h. 
Let Q approach P, so that h tends to zero. 
lin+ 
r lim -— 
Ax->o Ax 
lim (4\— h) 
A-f 0 
4 
The limit of the secant slopes is 4. Therefore the tangent slope is 4. 
When the slope of the tangent line through P is known, the equation 
of the tangent can be found from the point slope form of the equation 
of a line. For instance, the tangent line of Example 2 has equation 
y — y-i — m(x — xi) 
y + 3 = 4(x + 1) 
4x — y + 1 = 0 
52 2.2jSlope of the Tangent Line 
Example 8 
Find the slope of the tangent, and the equation of the tangent line, 
to the curve y = x3 at the origin. 
Solution 
Take P to be the origin (0, 0) and Q the point (h, h3) where h € R, 
h ^ 0. 
Ax = h — 0 
= h 
Ay = h3 — 0 
= h3 
Ay _ la_ 
Ax h 
= h2 
Now let Q approach P, so that h —> 0. 
• 1 ' 1 • 7 2 hm —— = Jim n 
4no Ax /,_»o 
Ax 
Hm -A = 0 
4j-»o Ax 
The limit of the secant slopes is zero. Therefore, the tangent slope at 
the origin is zero. 
The tangent line is the x axis, with equation y = 0. 
Note that the x axis crosses the curve y = x3 at the origin. Do 
any other lines tangent to the curve y = x3 cross the curve at the 
point of contact? 
When the tangent to a curve C at a point P has a given slope, 
we may say that the slope of the curve itself at P is equal to the slope 
of the tangent. Thus we make use of the fact that the tangent line 
to C has a unique slope number to define the slope of the curve at 
P. For instance, the slope of the curve y = x2 at (2, 4) is 4, and the 
slope of the curve y = x3 at the origin is zero. 
2.2/Slope of the Tangent Line 53 
Exercise 2.2 
1. Define the tangent to a curve at a given point. 
2. State which of the following sketches illustrate the given proper¬ 
ties (a), (b), (c), (d) of tangents to certain curves. 
(4) 
r^\ 
(a) The tangent meets the curve in more than one point. 
(b) The tangent crosses the curve. 
(c) The tangent does not cross the curve. 
(d) The tangent touches the curve at two points. 
3. Find the slope of the tangent to each of the given curves at the 
point with x coordinate 3. 
(a) y = x2 (b) y = x3 
4. Find the slope of the tangent to each of the given curves at the 
point with x coordinate —2. 
(a) y = x2 — 6x (b) y = x3 — 2x 
5. (a) Find the slopes of the tangents to the parabola y = x2 at the 
points with x coordinates —3, —1, 1, 3. 
(b) Graph the parabola and with a ruler draw tangents at the 
points listed in part (a). 
(c) Calculate from measurements the slopes of the tangents drawn 
in (b), and compare them with the slopes listed in (a). 
6. Repeat Question 5 for the parabola y = — x2 + x with tangents 
at points with x coordinates 1 and 3.7. 
8. 
Repeat Question 5 for the hyperbola y = - with tangents at 
x 
points with x coordinates 1 and 2. 
Repeat Question 5 for the curve y = x3, with tangents at points 
with x coordinates 1 and —2. 
9. Repeat Question 5 for each of the given curves, with tangents at 
points with x coordinates —1 and 1. 
(a) y = x3 — 3x (b) y = x3 + x2 
2.2/Slope of the Tangent Line 54 
2.3 Tangents to a Curve 
We have learned how to find the slopes of tangent lines to certain 
curves. We shall now find formulas for the slopes of the tangent at 
general points of such curves. It is then possible to study the way in 
which the slope of the tangent depends on the position of the point 
of contact on the curve. 
Example 1 
Find the slope of the tangent to the parabola y = x2 at the general 
point (xi, xi2). 
Solution 
Let P have coordinates (xi, Xi2), and 
let Q have x coordinate x2 = Xi + h, h ^ 0. 
Therefore Ax = x2 — Xi = h. 
The y coordinate of Q is 
x22 
(xi + h)2 
Xi2 + 2xi h + h2 
Xi2 
y2 — yi 
Xi2 + 2xih + h2 — xi2 
2xi h + h2 
2xi + h 
y2 = 
V y1 = 
Ay = 
Ay 
Ax 
Thus, the slope of the secant PQ is 2xi + h. 
The slope of the tangent is found by letting h approach zero. Thus, 
as h —► 0, 2xi + h —»2xi, 
Ay 
hm — = 2xi. 
Ax-+o Ax 
The limit of the slope of the secant is 2xx. Therefore the slope of the 
tangent to y = x2 at the point with x coordinate Xi is 2xi. 
In Example 1, 2x is an expression for the slope of the tangent at 
any point of the curve in terms of the x coordinate of that point. 
2.3/ Tangents to a Curve 55 
The graph of the curve, y = x2, with several tangents is shown in 
We have shown that the slope of the tangent to the parabola 
y = x2 at the point (x, x2) is 2x. When x is positive, the tangent 
slope is positive. Hence, to the right of the origin the tangent line and 
the curve slope upward. If x is positive and large, then the slope is 
also large, so that the tangent line and the curve are steep. 
Likewise, if x is negative, the slope is negative so that the curve 
slopes downward to the right. Further to the left the downward slope 
is steeper. 
Example 2 
Find the slope of the parabola y = 2x — x2 at the point (xi, y{) 
The slope of the parabola at a given point is defined as the slope 
of the tangent to the parabola at that point. 
Solution 
Let P be the general point (xi, ya) of the curve. Let Q be a second 
point (x2, yz) where X2 = xi + h, h 9^ 0, and therefore 
Ax = x2 — Xi = h. 
V X2 = Xi T h 
y2 = 2x2 — x22 
= 2(xi + h) — (xi + h)2 
= 2xi + 2 h — Xi2 — 2xi h — h2 
yi — 2xi — xi2 
A y = y2 — yi 
= 2h — 2xi h — h2 
= h(2 - 2xi - h) 
56 2.3/Tangents to a Curve 
Therefore, the slope of the secant PQ is 
^ = 2 - 2xx-h 
Ax 
To find the slope of the tangent let Q approach P. Thus, h-* 0 and 
the slope of the tangent is 
\y 
lim —— = lim (2 — 2xi — h) 
Therefore, the slope of the tangent to y = 2x — x2 at {xx,yi) is 
2 — 2xi = 2(1 - Xl). 
For what values of Xi is this slope 2(1 — Xi) positive? Zero? 
Negative? How are these facts related to the graph of the curve 
y — 2x — x-? The graph of the curve, y = 2x — x2, with tangents 
drawn at several points, is shown in Figure 2.2. 
Find the slope of the curve y = x3 at the point (xj, yi) on the curve. 
Solution 
Let P be the point (xi, yi) on the curve. Then yi = Xi3. Let Q be 
a point (x2, y-i) on the curve with x2 = Xi + h, so that h ^ 0. 
Ax = x2 — Xi 
= h 
And y2 = x23 
= (xi + h)3 v x2 = Xi + h 
= Xi3 + 3xi 2/z. + 3xi h2 + hz 
y i = xi3 
A y = y2 — yi 
= 3xi-Ji + 3xi h2 + hz 
2.3/ Pangents to a Curve 57 
The slope of the secant is 
^ = 3xi2 + 3*i h + h\ 
Ax 
Now let Q approach P, so that h—>0. Then, the tangent slope is 
lim ~ = lim (3xi2 + 3xih + h2) 
ino Ax /(—>o 
Ay 
that is, lim —— = 3xi2. 
Ax—>0 
The slope of the curve y = x3 at (xi, Xi3) is equal to 3xi2. 
For what values of Xi is the slope 3xi2 positive? Zero? Negative? 
The graph of the curve, with several tangents, is shown in Figure 2.3. 
For Xi £ R, 
xi2 > 0 
3xi2 > 0 
The slope is never negative. For Xi = 0 the slope is zero, but other¬ 
wise it is positive. Note that the curve rises steeply towards the right, 
but 1 as a “shelf” or “ledge” at the origin. 
Exercise 2.3 
1. If a portion of a curve rises toward the right, is its slope at any 
point of that portion positive or negative? 
2. If a portion of a curve rises steeply toward the left, is its slope at 
any point of that portion positive or negative? large or small? 
58 2.3/Tangents to a Curve 
3. Given the curve y = /(x), where /(x) = fx2. 
(a) Find the slope of the curve at a point (xi, y-d) on the curve. 
(b) Tabulate the coordinates and slopes for xx = —3, —2, — 1, 
0, 1, 2, 3. 
(c) Graph the tangent lines to the curve at xx = -3, — 1, 1, 3. 
(d) Sketch the graph of the curve on the diagram of (c). 
4. Given the curve y = g(x), where g(x) = —^x3. 
(a) Find the slope of the curve at a point (xj, yi) on the curve. 
(b) Tabulate the coordinates and slopes for xx = —3, —2, — 1, 
0, 1, 2, 3. 
(c) Graph the tangent lines to the curve at the points listed in (b). 
(d) Sketch the graph of the curve on the diagram of (c). 
5. Find the slope of the tangent line to the curve y = x4 at a point 
(xi, yi) on the curve. 
6. From your knowledge of the slopes of the tangents to the curves 
y = X2, y = X3, and y = x4, conjecture the slope of the tangent 
at (xi, yi) to the curve y = xn, n £ N. 
7. Find the points of contact of all tangents to the curve y = —x3 
that have the given slopes. 
(a) -3 (b) -12 (c) 0 (d) 1 
C 8. Show that the tangent to the curve y = x3 at (1, 1) also meets 
the curve at a second point. Find the slope of the tangent to 
the curve that touches the curve at the second point. 
9. (a) On the upper half of a sheet of paper draw a large smooth 
freehand curve similar to each of the given curves. 
(b) Mark five equally spaced points on your curve and draw 
tangents at these points with a ruler. 
(c) Using coordinate axes parallel to the edges of the paper, 
calculate the slopes of the tangents in (b). 
(d) Plot the slopes obtained in (c) using a set of axes situated 
below your sketch, on the lower half of the sheet. 
('e) Join the points plotted in (d) with a smooth freehand curve. 
2.3/ Tangents to a Curve 59 
2.4 The Derivative Function 
Let /: x—»/(x) be a function with domain R. Let us look for a 
formula or expression for the rate of change of / at x. Let P be a 
point on the curve y = fix) ; then P is (x, y) — (x,/(x)). Let Q have 
x coordinate x + h, so that Ax = h. Then, Q has y coordinate/(x + h), 
and 
Ay =/(x + h) - fix). 
The slope of the chord PQ is 
Ay _ f(x + h) - fix) 
Ax h 
y 
The rate of change of/ at x is the limit of this slope as Q approaches 
P, that is, as h = Ax —> 0, and so is denoted by 
lim ^ - lim /<* + hl ~ /(») . 
Az->0 Ax « 
We have already seen in several examples, the practical meaning 
of this symbol, and how to calculate the value which it denotes. 
It is convenient to have a shorter notation for the limiting rate 
of change. One such notation commonly used since the time of Leib¬ 
niz and Newton is the symbol 
dy 
dx 
This symbol is read “dee y by dee x,” and it is defined as the value 
of the given limit. The notation arises from the quotient — of the 
Ax 
two differences Ax and Ay. Whereas Ax and Ay denote numbers, the 
symbols dx or dy do not, by themselves, have any numerical values. 
The notation dy does suggest that in some sense dy is a limit of 
Ay as Ax tends to zero. But if the values zero were assigned to dx 
60 2.J)/ The Derivative Function 
and dy, the quotient ~ would be an undefined symbol —. You are 
dx 0 
cautioned, therefore, to remember that as yet the entire symbol — 
dx 
is the only quantity that has been given a meaning. 
dy 
lim ^ 
dx a^o Ax 
Example 1 
If y = x2, find . 
dx 
Solution 
By definition, 
(x + li)2 — x2 
lim--- 
h—>o n 
2 xh + h2 
lim--- 
h-> o n 
2x. 
dy