limits and continuity - ksuweb.kennesaw.eduksuweb.kennesaw.edu/~plaval/math1190/intro.pdf4 chapter...

Chapter 1

Limits and Continuity

1.1 Introduction

1.1.1 What is Calculus?

The origins of calculus can be traced back to ancient Greece. The ancient Greeksraised many questions about tangents, motion, area, the in�nitely small, thein�nitely large. The Greeks provided a few answers to their questions. But mostof them remained unanswered until the invention of modern calculus. After theGreeks, progress was very slow. Algebra, founded by Arab scholars in the ninthcentury was not fully systemized until the sixteenth century. In the seventeenthcentury, Descartes established analytic geometry. With algebra and analyticgeometry, the stage was set for calculus to evolve.The actual invention of modern calculus is credited to the Englishman Sir

Isaac Newton (1642-1727) and the German mathematician Gottfried WilhelmLeibniz (1646-1716). Newton started his work in 1665, Leibniz in 1673. Within100 years, calculus reached pretty much the state under which it is known today,though some theoretical subtleties were not fully resolved until the twentiethcentury.To a Roman during the Roman Empire, a "calculus" was a pebble used

in counting or gambling. The word then evolved to "calculare", which meant"to compute", to "reckon", "to �gure out". To a mathematician or scientist oftoday, calculus is elementary mathematics (algebra, geometry, trigonometry),enhanced by the limit process.Calculus is also the mathematics of motion and change. In other subjects

related to mathematics such as algebra, arithmetic, geometry, you have learnedprimarily to calculate with numbers and variables, to simplify algebraic ex-pressions, to deal with points, lines and �gures in the plane. But in each case,the quantities you dealt with were static. Calculus deals with quantities whichchange, quantities which approach other quantities. A technique often used incalculus when solving a di¢ cult problem is to look at a simpler problem; onewe know how to solve. We then change the simpler problem little by little. The

1

2 CHAPTER 1. LIMITS AND CONTINUITY

changes satisfy two criteria. The �rst is that as we apply the changes, the sim-pler problem looks more and more like the original, more di¢ cult problem. Thesecond criterion is that at each step, the changes should be such that we knowhow to solve the newly obtained problem. This technique illustrates the limitprocess mentioned above. We now look at some problems arising in Calculuswhich illustrate this technique.

1.1.2 Area of a Circle

It is easy to �nd the area of a circular region of radius r because we knowthe formula. It is �r2. How did people do it before the formula was known?One approach was to use inscribed regular polygons. For example, we couldapproximate the area of a circle by the area of the inscribed equilateral triangleas shown in �gure 1.1. Let A3 denote the area of the triangle. We have replacedthe problem of �nding the area of a circle by �nding the area of a triangle.Which is much easier to do. Of course, you should argue that the two areasare not the same. So, we see that the simpler problem is indeed easier to solve,however, it does not give us the correct answer. We don�t want to give up thatquickly though. Let us increase the number of sides of the polygon inscribed inthe circle. In general, let An denote the area of the regular polygon with n sidesinscribed in the circle. As �gures 1.2, 1.3, 1.4 and 1.5 suggest, the larger n is,the closer An is to the area of the circle.. We say that the area A of the circleis the limit of the areas of the inscribed polygons as n approaches in�nity, andwe write

A = limn!1

An

Of course, there is a slight problem. Even if we can derive a formula in termsof n for An, how do we compute lim

n!1An? Since 1 is not a number, we cannot

plug it in the formula.

Figure 1.1: Inscribed Triangle

1.1. INTRODUCTION 3

Figure 1.2: Inscribed square

Figure 1.3: Inscribed Pentagon

Figure 1.4: Inscribed Hexagon


Figure 1.5: Inscribed dodecagon

1.1.3 The Area Problem

The problem of �nding the area of a circle can be generalized to that of �ndingthe area of a region bounded by a curve. Consider the problem of �nding thearea S between the x-axis and the graph of y = f (x), between the vertical linesx = a and x = b. Such a region is shown in Figure 1.6. This region is toocomplex; there is not an easy formula which gives its area. We use the followingtechnique: We subdivide the interval [a; b] into n subinterval of equal length,and draw rectangles as shown in Figures 1.7, 1.8 and 1.9. Let Sn denote thesum of the areas of each rectangle. We approximate S by Sn. Obviously, forsmall n, like the case n = 2 shown in Figure 1.7, our approximation is not verygood. But, as n gets larger, Sn is closer and closer to S. Once again, using thenotation of the previous example, we have

S = limn!1

Sn

Of course, we have the same problem. Even if we can derive a formula in termsof n for Sn, how do we compute lim

n!1Sn? Since 1 is not a number, we cannot

plug it in the formula. This problem is related to the de�nite integral and willbe studied when we study integration. You can experiment with the techniquedescribed here using the applet at

http : ==science:kennesaw:edu=~plaval=applets=Riemann:html

Details of this procedure will be done in chapter 4.

1.1.4 The Tangent Problem

This is a classical problem in Calculus. We are going to look at it in greaterdetail. Before we start, let us review some essential formulas.

Proposition 1 The slopem of the line through two points of coordinates (x1; y1)and (x2; y2) is given by

m =y2 � y1x2 � x1

(1.1)

1.1. INTRODUCTION 5

Figure 1.6: Area of a Region

Figure 1.7: Two Rectangles


Figure 1.8: Four Rectangles

Figure 1.9: Eight Rectangles

1.1. INTRODUCTION 7

Proposition 2 The equation of a line through a point of coordinates (x1; y1)with slope m is

y � y1 = m (x� x1) (1.2)

If instead of being given the slope and a point we are given two points onthe line, we can �nd its equation by �rst using formula 1.1 to derive the slope ofthe line. Then, using the slope just derived, one of the given points and formula1.2, we can �nd the equation of the line.

Example 3 Find the equation of the line through the points (1;�1) and (2; 1).The slope of this line is

m =1� (�1)2� 1

=2

1= 2

The equation of the line is

y � (�1) = 2 (x� 1)y + 1 = 2x� 2

y = 2x� 3

We used the point (1;�1) to derive the equation. We could have also used thesecond point, we would have obtained the same result.

Now, we can look at the tangent problem. Suppose we are trying to �ndthe equation of the tangent line to the graph of y = f (x) at a point P on thegraph. Let (a; b) be the coordinates of P (since P is on the graph, we haveb = f (a)). For now, think of the tangent as the line which touches the graphat P . Such a line is shown in Figure 1.10.The tangent line is a line. To �nd theequation of a line, we either need two points on the line, or the slope of the lineand a point. Unfortunately, we do not have all the necessary information. Weare only given one point. Instead, we consider an easier problem, one we cansolve. The problem of �nding the equation of a line given two points. We usethe following approach:

1. Pick a second point Q on the graph, let (x; y) be its coordinates (note thatsince the point is on the graph, we have y = f (x)).

2. Find the slope of the secant line through P and Q, call it mPQ. Once wehave the slope, we can �nd the equation of that secant line. This secantline is shown in Figures 1.11 and 1.12. Of course, the secant line is notthe same as the tangent line. From the equations above, we see that

mPQ =f (x)� f (a)

x� a (1.3)


Figure 1.10: Tangent Line at P

3. Observe that by taking Q closer and closer to P , the secant line is gettingcloser and closer to the tangent line. In other words, the slope m of thetangent line through P can be though of as the limit of the slopes mPQ

of the secant lines through P and Q, as Q approaches P . Using the samenotation as above, we have

m = limQ!P

mPQ

To make Q closer to P is the same as making x closer to a. Thus,

m = limQ!P

mPQ (1.4)

= limx!a

mPQ (1.5)

= limx!a

f (x)� f (a)x� a (1.6)

Here again, we can see that this computation poses a problem becauseas x ! a, the denominator of this fraction approaches 0, which is notallowed.

This procedure is also illustrated by a Java applet which can be found at

http : ==science:kennesaw:edu=~plaval=tools=di�erentiation:html

In fact, the graphs shown in Figures 1.10, 1.11 and 1.12 come from this applet.The above procedure allows us to express the slope of the tangent. Once we

know how to compute limits, we will be able to �nd the slope of the tangent.

1.1. INTRODUCTION 9

Figure 1.11: Secant Line Through P and Q

Figure 1.12: Secant Line Through P and Q


Example 4 The purpose of this example is to illustrate the procedure outlinedabove. Consider the function f (x) = x2 + 5. Suppose we are trying to �nd theequation of the tangent at x = 3. To do so, we approximate it by �nding theequation of the secant line through x = 3 and another point we will call x = a.Then, we let a get closer and closer to 3. This amounts to using values of awhich are closer and closer to 3. Answer the questions below:

1. Find the equation of the secant line through the points corresponding tox = 3 and x = 4.Note that the two points in questions are P = (3; f (3)) and Q = (4; f (4))that is (3; 14) and (4; 21). Using formula 1.3, we see that the slope of thesecant line is

mPQ =f (4)� f (3)

4� 3

=21� 141

= 7

Thus, the equation of the secant line (using the point (3; 14) is

y � 14 = 7 (x� 3)y � 14 = 7x� 21

ory = 7x� 7

2. Find the slope of the secant line through the points corresponding to x =3and x = a, for a = 4, 3:5, 3:1, 3:01, 3:001.We proceed as above. The slope of the secant line is

mPQ =f (a)� f (3)

a� 3

=a2 + 5� 14a� 3

=a2 � 9a� 3

=(a� 3) (a+ 3)

a� 3= a+ 3 as long as a 6= 3

The table below gives us the slope of the secant for the required values ofa.

a 4 3:5 3:1 3:01 3:01mPQ 7 6:5 6:1 6:01 6:001

1.1. INTRODUCTION 11

3. Write a formula for the slope of the tangent to y = f (x) at x = 3.If we call m the slope of the tangent, using formula 1.6, we have

m = limx!3

f (x)� f (3)x� 3

= limx!3

x2 + 5� 14x� 3

= limx!3

x2 � 9x� 3

1.1.5 The Velocity Problem

Velocity is de�ned to be

velocity =distancetime

For example if you drive 100 miles in 2 hours, then your velocity was 50miles/hour. This is known as the average velocity, that is the velocity overa fairly large time interval. It does not mean your velocity was 50 miles/hourduring the whole trip. Indeed, if you looked at the speedometer of your car,you probably notice that it was changing all the time. This is because thespeedometer measures the velocity at every instant. It measures what we callthe instantaneous velocity. We illustrate these two notions.Suppose that the position of an object is given by a function s (t). For

example, if t is in seconds and s (t) in meters, then s (2) represents the distancein meters traveled by the object after 2 seconds, s (5) represents the distancetraveled after 5 seconds, and so on. Therefore, s (b)�s (a) represents the distancetraveled between t = a and t = b. So, we have

De�nition 5 (Average Velocity) If the position of an object is given by thefunction s (t) then the average velocity of the object between t = a and t = b isgiven by

average velocity =s (b)� s (a)b� a (1.7)

Remark 6 Geometrically, the average velocity represents the slope of the secantto y = s (t) through the points corresponding to t = a and t = b. So, we see thatthe velocity problem is identical to the secant problem.

To �nd the instantaneous velocity, we proceed the same way we did for thetangent problem. Since an instantaneous velocity is an average velocity overa very small interval, we can approximate the instantaneous velocity at t = aby computing the average velocity between a and t, then we let t approach a,thus making the time interval smaller and smaller. In other words, we have thefollowing:

De�nition 7 (Instantaneous Velocity) If the position of an object is givenby the function s (t) then the instantaneous velocity of the object at time t = ais


given by

v = limt!a

s (t)� s (a)t� a (1.8)

Remark 8 Geometrically, the instantaneous velocity at t = a is the slope ofthe tangent to the graph of y = s (t) at t = a.

We illustrate this with an example.

Example 9 When an object fall, the distance (in meters) it travels is given bys (t) = 4:9t2 where t is in seconds.

1. Find the average velocity of the object between t = 1 and t = 2.From formula 1.7, we have

average velocity =s (2)� s (1)2� 1

=19:6� 4:9

1= 14:7 m=s

2. Express the instantaneous velocity of the object at t = 1 as a formula.From formula 1.8, we have

v = limt!1

s (t)� s (1)t� 1

= limt!1

4:9t2 � 4:9t� 1

= limt!1

4:9�t2 � 1

�t� 1

1.1.6 Conclusion

When doing Calculus, like in many other disciplines, it is important to knowthe details. It is also important to know the general idea. We only presenteda few problems here in which the same idea, the idea of �nding a quantity asthe limit of other quantities, is used. This idea is central to Calculus. We willlook at all the examples presented above in more details. However, before wedo this, we must explain in more details this idea of limits.

1.1.7 A Note on Functions

It is assumed that the reader is already familiar with the notion of functions.The functions encountered in this class fall in two categories. They are either anelementary function or a combination of elementary functions. Given a function,understanding its pattern, that is how it is made, is essential to understandingthe properties of this function and to knowing how to work with it. We review


both the elementary functions and the ways to combine them. In what follows,C will denote a constant, n will denote a positive integer constant, a will denotea positive real number constant not equal to 1, f and g will denote two functionsand x will denote the independent variable.

1. The elementary functions the reader will encounter in this class are:

(a) C (constant function)

(b) xn (power function)

(c) npx (nth root function)

(d) ex and ax (exponential functions)

(e) lnx and loga x (logarithmic functions)

(f) sinx, cosx, tanx, cotx, secx, cscx (trigonometric functions)

(g) sin�1 x or arcsinx, cos�1 x or arccosx, tan�1 x or arctanx (inverseof the trigonometric functions)

2. Functions can be combined as follows:

(a) f + g where (f + g) (x) = f (x) + g (x) (addition)

(b) f � g where (f � g) (x) = f (x)� g (x) (subtraction)(c) Cf where (Cf) (x) = Cf (x) (multiplication by a constant)

(d) fg where (fg) (x) = f (x) g (x) (multiplication of two functions)

(e)f

gwhere

�f

g

�(x) =

f (x)

g (x)(division of two functions)

(f) f � g where (f � g) (x) = f (g (x)) (composition of two functions)

When trying to decide if a given function has a certain property, this willdepend on what the function looks like. If the function is one of the elemen-tary functions, then it is simply a matter of knowing whether that elementaryfunction has the given property. However, if the given function is a combina-tion of elementary functions, then one needs to see if each elementary functionhas the property as well as whether that property is preserved by the way theelementary functions are combined. We illustrate this with a few examples.

Example 10 What is the domain of f (x) = (x� 1) lnx?The property in question here is "being de�ned". We are asking where f (x) isde�ned. First, we note that f is a product of two functions lnx and x� 1. Weknow that a product is always de�ned as long as each member of the product isde�ned. Since x � 1 is always de�ned and lnx is de�ned only when x > 0, weconclude that f (x) is de�ned when x > 0 that is on (0;1).


Example 11 What is the domain of g (x) =lnx

(x� 1)?

The property in question here is "being de�ned". We are asking where g (x)is de�ned. First, we note that g is a division of two functions lnx and x � 1.For the division to be de�ned, each function has to be de�ned. In addition, thedenominator cannot be 0. We see that in this case, the division adds an extracondition. From the previos example, we know that x � 1 is always de�ned.lnx is de�ned when x > 0. Since the denominator cannot be 0, it follows thatx� 1 6= 0 or x 6= 1. In conclusion, g (x) is de�ned when x > 0 and x 6= 1.

Remark 12 The last example shows that a property depends not only on eachelementary functions making up a given function, but also on how the elementaryfunctions are combined.

The technique used in �nding the domain in the two examples above will beused throughout Calculus when deciding if a function has a certain property orwhen applying a certain rule to a function. It will always involve understandinghow a given function is made up. Which elementary functions it contains andhow they are combined.

1.1.8 Sample Problems

The �rst three problems are related to the material of this section. Problem 4is a review problem. In the next few sections, the concept of the domain of afunction will play an important role. These problems are designed to check thatstudents have mastered this concept. If it is not the case, students are stronglyencouraged to review this concept. The remaining problems are designed toprepare students for the material to come.

1. Consider the function f (x) = x2 � 3

(a) Find the equation of the secant line through the points correspondingto x = 1 and x = 2. (answer y = 3x� 5)

(b) Write a formula for the slope of the tangent at x = 1.�answer lim

x!1

(x� 1) (x+ 1)x� 1

�2. Consider the function f (x) = sinx

(a) Find the equation of the secant line through the points corresponding

to x = 0 and x =�

2.�answer y =

2

�x

�(b) Write a formula for the slope of the tangent at x =

�

2.�answer lim

x7�!�2

sinx� 1x� �

2

�3. The position of an object in feet is given by s (t) = 40t� 16t2.

(a) Find the average velocity of the object between t = 1 and t = 2.(answer � 8 ft= sec)


(b) Express the instantaneous velocity of the object at time t = 1 as a

formula.�answer lim

t7�!1

�8 (t� 1) (2t� 3)t� 1

�4. Find the domain of the functions below

(a) ln (x� 2). (answer (2;1))(b)

p(x� 2). (answer [2;1))

(c) ln (5� x). (answer (�1; 5))

(d)lnx

x� 5 . (answer (0; 5) [ (5;1))

(e)x� 5lnx

. (answer (0; 1) [ (1;1))

5. Consider a function y = f (x). What can you say about the output values(y values) corresponding to input values which would be very close toone another? You may look at speci�c examples to help you answer thisquestion.

6. Can you think of a function y = f (x) for which f (a) and f (b) would befar apart, even though a and b are very close? First, think what the graphof that function would look like. Then, try to �nd a formula for it.

limits and continuity - ksuweb.kennesaw.eduksuweb.kennesaw.edu/~plaval/math1190/intro.pdf4 chapter...

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