2

Calculus

About this book : http://matrix.skku.ac.kr/Cal-Book/

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Chapter 2. Limits and Continuity

2.1 Limits of functions

[Solutions-Presentation] http://youtu.be/LZSmRPAAXME

[Solutions-Presentation] http://youtu.be/hj8d-j_DGf4

2.2 Continuity http://youtu.be/zGxx3PUCTnM

[Solutions-Presentation] http://youtu.be/azrkT1RP4-c

2.1 Limits of Functions

[lectures] http://youtu.be/VBCeAllP1M0

[Solution] http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-2-1-Sol.html

In this section, we introduce the notion of the “limit.” Intuitively, this notion is easy to understand. But mathematically, we should prepare for situations that are not intuitive at all. As the chapter progresses we will introduce a more rigorous definition of the limit. Thus, for completeness, we will define the “limit” by ‘-’ arguments at the end of this section. We also will study many limit laws and most of them can be proved rigorously with ‘-’ arguments.

First, consider one example

EXAMPLE 1 Investigate the behavior of the function , , near the point .

Solution. Note that is not defined because the denominator is zero when Compute some values of when is close to 2, and smaller than 2. (We say “from the left” of 2.)

1.9

1.99

1.999

1.9999

1.99999

1.999999

3.9

3.99

3.999

3.9999

3.99999

3.999999

From the above table, we observe that as is getting closer and closer to 2 from the left (), the function value is getting closer and closer to 4.

On the other hand, compute some other values of when is close to 2 and larger than 2 from the right.

2.1

2.01

2.001

2.0001

2.00001

2.000001

4.1

4.01

4.001

4.0001

4.00001

4.000001

From the above table, we observe that as is getting closer and closer to 2 from the right (), the function value is also getting closer and closer to 4 but not equal to 4. So if approaches 2, then we have only two cases, that is, (i) and approaches 2, or (ii) and approaches 2. From above arguments, we find that the function value approaches 4 in both cases. ■

The above example suggests the following definition of the limit.

DEFINITION 1

The expression “” is read as “the limit of , as approaches , equals .” This means that we can make the values of to be arbitrarily close to by taking to be sufficiently close to but not equal to . In other words, gets closer and closer to as gets closer and closer to (but not equal to ).

Alternatively, we write ‘ as .’

In finding the limit of as approaches , the phrase “but ” means that is never considered. Thus we do not care about the value of for finding the limit of . It does not matter even if is not defined at . Our interest is in how is defined near . Figure 1 shows the graphs of three functions. Note that in part (b), , in part (c), is not defined. But in each case, regardless of what happens to at , it is true that .

Figure 1 three cases

DEFINITION 2

The expression “” is read as “the left-hand limit of , as approaches , equals ”. This means that we can make the values of arbitrarily close to by taking to be sufficiently close to and is less than . In other words, gets closer and closer to as gets closer and closer to with restriction .

Similarly, if we restrict that to be greater than , then we have “the right-hand limit of as approaches , equals .” For notation, we use

Thus, the symbol “” means that we consider only (only values of that are greater than , that is, to the right of are considered). Similarly “” means that we consider only (only values of that are less than , that is, to the left of are considered).

In the Figure 1, the limit of exists if the limits of ‘from the right and left’ coincide. We will call the limit from the left and from the right, ‘left-hand limit’ and ‘right-hand limit’, respectively.

We have the following theorem.

THEOREM 3

The limit of as approaches exists if and only if both left and right limits as approaches exist and coincide. That is,

if and only if

In some cases, both the left-hand and the right-hand limits exist, but they are not equal, so the limit does not exist. Consider the Heaviside function : (The Heaviside function is very important in ordinary differential equations and electromagnetism because this function represents turning on a switch at time 0).

For the Heaviside function, we have and Hence does not exist.

EXAMPLE 2

Determine whether the limit exists or not.

Figure 2

Solution.

Here we use the Theorem 3 above. That is, we consider the left -hand and the right-hand limits. If we consider the left-hand limit, then we have , because . Thus we have the left-hand limit On the other hand, if we consider the right-hand limit, then we have , because . Hence we have the right-hand limit . Because these two limits do not coincide, does not exist by Theorem 3. ■

Here we describe what it means for the values of a function to increase (or decrease) without bound or to approach infinity and what it means for to have a limit as approaches infinity . The symbol does not represent any real number. Also we cannot use in arithmetic in the usual way.

Infinite Limits

The following definition of infinite limits is informal, but it is adequate for most functions encountered in this book.

DEFINITION 4

Suppose is defined for all near . If grows arbitrarily large for all sufficiently close(but not equal) to , we write

We say the limit of as approaches is infinity.

If is negative and grows arbitrarily large in magnitude for all sufficiently close(but not equal) to , we write

In this case, we say the limit of as approaches is negative infinity. In both cases, the limit does not exist.

For example consider the function which is defined for all real numbers except at .

Let us consider the limit of , and observe that:

(ⅰ) approaches 0 as tends to .

(ⅱ) tends to as approaches 0 from the right.

(ⅲ) tends to as approaches 0 from the left.

(ⅳ) approaches 0 as tends to .

Let a function be defined on both sides of , except possibly at itself. Then means that the values of can be made arbitrarily large (as large as we want) (or “increase without bound”) by taking sufficiently close to , but of course not equal to . In this case we say that the limit does not exist.

Figure 3

Similarly, means that the values of can be made arbitrarily large and negative by taking sufficiently close to , but not equal to .

Similar definitions can be given for the following one-sided limits

, , , .

Consider the natural logarithm function . (See Figure 3.) As , tends to and the distance between curve and -axis approaches 0.

We will call the straight line the vertical asymptote of .

THEOREM 5

A straight line is said to be an asymptote to a curve if the distance between the curve and the line decreases to zero as we traverse along the curve . Asymptotes may be horizontal, or vertical.

If (a finite number) as or , then is a horizontal asymptote.

Figure 4

Similarly if (or ) as (finite number), then is a vertical asymptote. The calculation of slant asymptotes is more involved. In general, a knowledge of asymptotes is very useful when sketching graphs.

EXAMPLE 3

Find vertical and horizontal asymptotes of .

Solution. Now . So are two vertical asymptotes. Similarly are two horizontal asymptotes. ■

For calculations of limits, the following limit laws are very useful. Proofs of the limit laws can be constructed using - arguments.

Properties of Limit

Suppose that the limits and exist and is a constant. Then

5. if

6. , where is a positive integer.

7. , where is a positive integer.

(If is even, we require .)

http://www.milefoot.com/math/calculus/limits/GenericLimitLawProofs04.htm

Note: The Limit Laws also hold for one-sided limits.

Next, we state the Squeeze Theorem (or Sandwich Theorem), which is useful in finding many important limits.

THEOREM 6 The Squeeze Theorem (or Sandwich Theorem)

If when is near (except possibly at ) and

　 then

This says that if is squeezed between and near , and if and have the same limit at , then is forced to have the same limit at . (See Figure 5 and also imagine a sandwich.)

Figure 5 Squeeze Theorem

EXAMPLE 4

Find .

Figure 6

Solution. Since does not exist (the functions oscillates between and ), the fourth property of limit cannot be applied here.

However, since , we have . (See Figure 6.)

But we know that . Take , , , then by the Squeeze Theorem, we have . ■

CAS EXAMPLE 5

Find numerically by filling out the following form using the given Sage code.

http://matrix.skku.ac.kr/cal-lab/cal-2-1-Exm5.html

Solution.

var('x')

f(x)=(x^3-8)/(x-2)

a=[1.9, 1.99, 1.999, 1.9999, 1.99999, 1.999999, 1.9999999]

b=[2.1, 2.01, 2.001, 2.0001, 2.00001, 2.000001, 2.0000001]

print [f(k) for k in a]

print [f(k) for k in b]

1.9

1.99

1.999

1.9999

1.99999

1.999999

1.9999999

2.1

2.01

2.001

2.0001

2.00001

2.000001

2.0000001

[11.4100000000000, 11.9401000000000, 11.9940010000001,

11.9994000099957, 11.9999400000839, 11.9999940003548, 11.9999993960387]

[12.6100000000000, 12.0601000000000, 12.0060009999991,

12.0006000099972, 12.0000600001826, 12.0000060005334, 12.0000006039613]

Answer : It converges to 12. ■

[Practice Math & Code in http://sage.skku.edu/ or https://sagecell.sagemath.org/ or https://cocalc.com/]

Until now, we have used an intuitive definition of the limit. But if we want to prove the previous theorems and limit laws with mathematically rigorous deduction, we need the following “-” definition for the limit.

DEFINITION 7 Epsilon-Delta Definition

http://mathworld.wolfram.com/Epsilon-DeltaDefinition.html

An epsilon-delta definition is a mathematical definition in which a statement on a real function of one variable having, for example, the form "for all neighborhoods of there is a neighborhood of such that, whenever , then " is rephrased as "for all there is such that, whenever , then ." These two statements are equivalent formulations of the definition of the limit . In the second one, the neighborhood is replaced by the open interval , and the neighborhood by the open interval .

For a function of several variables, the absolute value would be replaced by the norm of . These statements express the fact that for all which lie close enough to , lies as close to as desired. This facilitates the task of proving limits since the fundamental formula are actually shown by constructing a with the required property for any given .

Let us prove the sum law using the above definition.

Limit Law (Sum Law)

If and both exist, then .　

Proof

The Sum Law Let be given. We must find such that whenever . 　

Using the Triangle Inequality we have

We make less than by making each of the terms and less than , respectively.

Since and there exists a number such that

whenever .

Similarly, since there exists a number such that whenever .　Let . Notice that if then and ,　and so and .　

Therefore, .　

To summarize, whenever .Thus, by the definition of a limit, ■

Figure 7 “-” definition

Consider the function if and (see Figure 7).

When is close to 4 but , it is clear from the definition that is close to 13, and so .

If should differ from 13 by less than 0.l, determine how much should differ from 4. Since the distance from to 4 is and the distance from from 13 is , the problem is to find a number such that if but , then .

By the side calculation, gives as a possible candidate of .

Notice that if then . Thus, . In other words, if is within a distance of from 4, then will be within a distance of 0.1 from 13. Similarly for , we get , for , we get and so on. The numbers , 0.01, and 0.001 are known as the error tolerances. Thus for any arbitrary positive number , there exists a positive number , namely , such that

[1] , if .

Note that [1] can be rewritten as whenever , () (see Figure 7). By taking the values of to lie in the interval , we can make the values of lie in the interval .

A precise definition of a limit can be given using [1] as a model. Let be a function defined on some open interval that contains the number , except possibly at itself. Then we say that the limit of is as approaches , and we write .

If, for every number , there is a number such that whenever 　 or if , then .

In other words, 　means that the distance between and can be made arbitrarily small by taking the distance from to sufficiently small (but not 0).　

Alternatively, means that the values of can be made as close as to by taking close enough to (but not equal to ).

Therefore, in terms of intervals, it can be stated as follows:

means that for every (no matter how small is) we can find such that if lies in the open interval and then lies in the open interval .

Let the function map a subset of ℝ onto another subset of ℝ (see the arrow diagram in Figure 8).

Figure 8

The definition of limit says that if any small interval is given around , then we can find an interval around a such that maps all the points in (except possibly ) into the interval . (See Figure 9.)

Figure 9

Figure 10

Another geometric interpretation of limits can be given in terms of the graph of a function. If is given, then we draw the horizontal lines and and the graph of (see Figure 10). If , then we can find a number such that if we restrict to lie in the interval but not equal to a, then the curve lies between the lines and . (See Figure 11.)

If a smaller is chosen, then a smaller may be required. (See Figure 12.)

Figure 11 Figure 12

EXAMPLE 6

Use the above definition to prove that .

Solution. Let be a given positive number. Find a number such that whenever . Since the choice of depends on , we can establish a connection between the absolute values and .

Consider . So for a given , choose . For this choice, if , then .

Therefore, by the definition of a limit we have . ■

For a one-sided limit, we have a similar - definition.

EXAMPLE 7

Prove that .

Solution. Let be a given positive number. Here and . To find a number such that whenever . Squaring both sides of the inequality , we get whenever .　This suggests us to choose . ■

(Movie on ‘-’ : http://www.youtube.com/watch?v=-ejyeII0i5c)

http://matrix.skku.ac.kr/cal-lab/SKKU-Cell-Epsilon-Delta.html

Precise Definition of Infinite Limits

Let be a function defined on some open interval that contains the number , except possibly at itself. Then means that for every positive number there is a positive number such that whenever .

So the values of can be made arbitrarily large (larger than any given number ) by taking close enough to (within a distance where depends on , but with ). (See Figure 13.)

Given any horizontal line , we can find a number such that when is restricted to lie in the interval but , then the curve lies above the line . If another , lager value, is chosen, a smaller may be required.

Figure 13

EXAMPLE 8

Show that .

Solution. Given any (large) number , find such that whenever . Since both and are positive, whenever . Taking the square root of both sides and recalling that , we get whenever .

So for any , choose . Now if , then , that is, . Thus whenever .

Therefore, . ■

CAS EXAMPLE 9

Find the following limit or explain why the limit does not exist. Solution.

var('x')

plot(cos(1/abs(x)), x, -1,1)

limit(cos(1/abs(x)), x=0)

Answer: Diverge, its limit does not exist.

Figure 14 ■

EXAMPLE 10

Plot the graph of the function near and check if the left-hand limit, right-hand limit and limit at exists.

Solution.

f(x) = sin(x)/(2*abs(sin(x)))

f.plot(exclude=[0])

f.limit(x=0, dir=‘+’)

f.limit(x=0, dir=‘-’)

f.limit(x=0) # x |--> undifined

Answer : Diverge, its limit does not exist.

Figure 15

■

Figure 15

2.1 EXERCISES (Limits of Functions)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-2-1-Sol.html

http://youtu.be/LZSmRPAAXME

1-4. Find the following limits or explain why the limit does not exist.

Solution.

Since , and , so .

Then does not exist.

CAS 3.

http://matrix.skku.ac.kr/cal-lab/cal-2-1-3.html

Solution. We draw with the following Sage command.

We see it diverge () as .

var('x')

p1=plot(sin(x)/abs(cos(x)), x, -pi,pi/2, color='blue');

p2=plot(sin(x)/abs(cos(x)), x, pi/2,pi, color='red');

show(p1+p2, ymax=50, ymin=-10)

We can find with the following Sage command.

limit(sin(x)/abs(cos(x)), x=pi/2, dir='plus')

Answer : +Infinity

Solution. Since when , .

5. http://matrix.skku.ac.kr/cal-lab/cal-2-1-5.html

Solution.

6. http://matrix.skku.ac.kr/cal-lab/cal-2-1-6.html

Solution.

P1=plot(1/abs(x)-1/x, x, -5,-0.1, fontsize=20, color='blue')

P2=plot(1/abs(x)-1/x, x, 0.1,2.5, fontsize=20, color='red')

CAS 7. http://matrix.skku.ac.kr/cal-lab/cal-2-1-7.html

Solution. We draw with the following Sage command.

We see the function converge to as .

p1=plot(x*sin(3/x), x, -0.5, 0, color='blue')

p2=plot(x*sin(3/x), x, 0, 0.5, color='red');

show(p1+p2, ymax=0.3, ymin=-0.3, aspect_ratio=1)

We can find with the following Sage command.

limit(x*sin(3/x), x=0) # 0

8. The sign function, denoted by , is defined by the following piecewise function.

Find the following limits or explain why the limit does not exist.

Solution. (a) (b) does not exist.

( )

9. Consider the function .

(a) Find and .

(b) Find the asymptotes of ; vertical and horizontal.

Solution. (a)

(b) : vertical asymptote (c) : horizontal asymptote

var('x, y')

p1=plot((x^2+2*x)/(x^2+x-2), x, -10, 10, color='blue');

show(p1, aspect_ratio=1, ymax=10, ymin=-10)

10. Draw the graph of a function with all of the following properties: (a) its domain is

(b) there is a vertical asymptote at

(c)

(d)

(e) does not exist.

(f) does not exist.

(g)

var('x')

f(x)=(-x^2 + 21)/(x + 3)^2 + 2

p=plot(f(x), (x, -10, 0), ymax=100)

pl=plot(f(x), (x, -10, 0), ymax=100, color="blue")

p2=plot(x^2 -6 , (x, 0, 3), ymax=100, color="blue")

p3=plot((-x-3)/(x-5), (x, 3, 5), ymax=100, color="blue")

g=p+pl+p2+p3

g.show()

11. Let .

(a) Find or explain why the limit does not exist.

(b) Find and such that for all .

Solution. (a) .

(b) Since , we have . Take, and .

CAS 12.Use the Squeeze Theorem to find .

http://matrix.skku.ac.kr/cal-lab/cal-2-1-12.html

Solution.

P1=plot(sqrt(5*x)*(4-cos(3/sqrt(x))), 0.01, 2)

P2=plot(sqrt(5*x)*(5), 0.01, 2, color='red')

P3=plot(sqrt(5*x)*(3), 0.01, 2, color='red')

show(P1+P2+P3)

CAS 13. Use the Squeeze Theorem to find .

http://matrix.skku.ac.kr/cal-lab/cal-2-1-13.html

Solution.

P1=plot(sin(x)/x, -4, 4, color='red')

P2=plot(cos(x), -4, 4, color='green')

P3=plot(1, -4, 4, color='blue')

show(P1+P2+P3)

(※ is not always smaller than , but it is when near 0. So the Squeeze Theorem can be used.)

14. Let . Find all positive integer such that .

Solution.

(ⅰ) ;

(ⅱ) , ( is positive integer) ;

(ⅲ) , ( is positive integer) ;

(ⅳ) ;

Therefore

var('x, y')

p1=plot((5*x^6-3*x^3+11)/(3*x^n-7*x+14), x, -5, 5, color='blue');

show(p1, aspect_ratio=1, ymax=10, ymin=-1

When .

15. Find all the asymptotes (vertical and horizontal) of the function .

http://matrix.skku.ac.kr/cal-lab/cal-2-1-15.html

Solution.

var('x');

show(plot(x^3/(x^2-x-2), x, -5, 5, color='blue'), ymax=5, ymin=-5)

, so ,,

and . Thus, and are vertical asymptotes.

16. Find the limit .

Solution.

17. Consider .

(a) Find all the vertical asymptotes for .

(b) If we restrict the domain to , then show that there exists an inverse function defined on .

Solution. (a) , so.

Thus, are vertical asymptotes.

18. Find such that whenever .

Solution.

19. Use an - argument to prove that .

Solution. Let be a given positive number. Here and . We must find a number such that whenever . With easy computation, we may choose to get the desired result.

20.Use the - argument to prove that if .

Solution. Let be a given positive number. We must find a number such that whenever . With an easy computation, we may choose to get the desired result.

21-26. Prove the statements using an - argument.

21.

Solution. Find . Let . If , then .

22.

Solution. Let be a given positive number. Here and . We must find a number such that whenever . With an easy computation, we may choose to get the desired result.

23.

Solution. Find . Let . If , then .

24.

Solution. Find . Let . If , then .

25.

Solution. Find . Let {}.

If , then =.

26.

Solution. Find . Let .

If , then

Take .

If , then .

27.

http://matrix.skku.ac.kr/cal-lab/cal-2-1-27.html

Solution. Find . Let

. If , then

and

. .

28.

Solution. Given any (large) number we must find such that whenever . Since both and are positive, whenever . Taking the square root of both sides and recalling that , we get whenever .

So for any , choose .

Now if , then , that is,

Thus whenever .

Therefore, .

29.

http://matrix.skku.ac.kr/cal-lab/cal-2-1-28.html

Solution. , Find . Let ().

If , then .

Since ,

30.

Solution. Let be a given positive number. Here and . We must find a number such that whenever . With an easy computation, we may choose to get the desired result.

31. Use an - argument to prove that

Solution. Given any (large) number we must find such that whenever . Since both and are positive, whenever .

So for any , choose . Now if , then .

Thus,

32. If and , where is a real number. Show that

(a)

(b) if

Solution. (a)

(b)

2.2 Continuity

[Lecture] http://youtu.be/zGxx3PUCTnM

A function is continuous at if . So if a function is continuous at , then will satisfy the following three conditions :

1. is defined (that is, is in the domain of )

2. exists

Figure 11

In other words, is continuous at if approaches as approaches (see Figure 1). Thus, a continuous function has the property that a small change in produces a small change in the corresponding value . Intuitively the graph of a continuous function has no break. So the graph can be drawn without raising the pen off the paper.

A function is said to be discontinuous at , or has a discontinuity at , if is not continuous . This can happen in three possible ways shown in Figure 2.

Thus, a function has a discontinuity at if one of the following three possible scenarios occurs:

ⅰ) is not defined at ;

ⅱ) is defined at but has no limit as ;

ⅲ) is defined at and exists, but is not equal .

Discontinuities fall into two categories: removable and non-removable. A discontinuity at is called removable if can be made continuous by defining (as in Figure 2(a)) or redefining at (as in Figure 2(c)). The discontinuity in Figure 2(b) and Figure 2(d) is non-remov

(a) (b) (c) (d)

Figure 2 Graphs of the functions in example 1

EXAMPLE 1

Classify the points of discontinuity of the following functions (see Figure 2).

(a) (b)

Solution. (a)It is easy to check that . Since is not defined so has a removable discontinuity at .

(b)At has a discontinuity because does not exist, although is defined.

exists. But so has a removable discontinuity at .

(d)The greatest integer function has jump discon- tinuities (“jumps” from one value to another) at all of the integers because does not exist if is an integer. ■

A function is continuous from the right at if and is continuous from the left at if .

For example, the function is continuous from the right but discontinuous from the left at each integer because but

A function is continuous on an interval if it is continuous at every number in the interval. (If is continuous at the endpoint it means continuous from the right or continuous from the left of the endpoint.)

THEOREM 3

If and are continuous at and is a constant, then the

following functions are also continuous at :

1. 2. 3.

4. 5. if

Proof Each of the five parts of this theorem follows from the corresponding Limit Law. For instance, we give the proof of part 1. Since and are continuous at , we have and . Therefore,

. ■

This shows that is continuous at .

THEOREM 2

(a) Any polynomial is continuous on .

(b) Any rational function is continuous on its domain.

Proof (a) Let the polynomial be a function of the form

where , , , are constants. We know that and (). Thus the function is a continuous function, and so is the function . Therefore, is continuous because is a sum of continuous functions.

(b) A rational function is a function of the form where and are polynomials. The domain of is . It follows from (a) that and are continuous everywhere. It follows from Limit Law 5, that is continuous at every number in . ■

Note: The inverse function of any continuous function is also continuous. (The graph of is obtained by reflecting the graph of about the line . So if the graph of has no break in it, neither does the graph of ).

EXAMPLE 2

Evaluate .

Solution. The function , is rational, so it is continuous on its domain, which is . Therefore, . ■

THEOREM 3

The polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions, are continuous on their domains.

EXAMPLE 3

Find where the function is continuous.

Solution. Since the function is continuous for and is continuous on ℝ, so is continuous on . The denominator, , is a polynomial, so it is continuous on . Therefore, is continuous at all positive numbers except where . So is continuous on the intervals or . ■

THEOREM 4 Limit of Continuous Functions

If is continuous at and , then .

In other words, .

Thus if the function is continuous and the limit of its argument exists then a limit symbol can be “moved through the function symbol”.

Consider . Then and . Thus (Limit Law 7), .

EXAMPLE 4

Find .

Solution. Since is a continuous function, using Theorem 3 we have

. ■

THEOREM 5 Composite of Continuous Functions

If is continuous at and is continuous at , then the composite function given by is continuous at .

In other words “the composition of continuous functions is also a continuous function.”

Proof Since is continuous at , we have . Since is continuous at , we to obtain . This means that the function is continuous at . Thus, is continuous at . ■

EXAMPLE 5

Determine where the functions

(i)

(ii) are continuous.

Solution.

Figure 3

(i) With , and we have . Since is a polynomial it is continuous on , the exponential function is continuous on , and is also continuous on . Thus is continuous on .

(ii) We know that is continuous and is continuous (because both and are continuous). Therefore, is continuous Figure 3 wherever it is defined. Now is defined when . It is undefined when , and this happens when , , .... Thus, has discontinuities when is an odd multiple of and is continuous on the intervals between these values. (See Figure 3.)

THEOREM 6 The Intermediate Value Theorem

Suppose that is continuous on the closed interval and let be any number between and , where . Then there exists a number in such that .

Thus a continuous function on must take on every intermediate value between the function values and at least once. (See Figure 4.)

(a) (b)

Figure 4 Intermediate Value Theorem illustrated

Geometrically, the graph of a continuous function cannot skip over any horizontal line given between and . It must intersect .

Observations:

(ⅰ) The value can be taken on once [as in Figure 4 part (a)] or more than once [as in part (b)].

(ⅱ)The Intermediate Value Theorem is not true in general for discontinuous functions.

(ⅲ)The Intermediate Value Theorem is useful in locating roots of equations (see corollary below).

The following theorem is an immediate and useful application of the Intermediate Value Theorem.

THEOREM 7 The Intermediate Value Theorem

If is continuous on and and have opposite signs , ], then there exists at least one number such that . (Recall that is known as a zero or root of .) (See Figure 5.)

Figure 6

Observe that this corollary is simply the special case of the Intermediate Value Theorem where .

The Intermediate Value Theorem and the above corollary are classical examples of existence theorems which tell the existence of a number , but do not specify what is.

EXAMPLE 6

Locate the zeros of .

Solution. Let us find the values of for several values of as follows.

−10

−5

−3

−2

−1

0

1

2

3

5

10

−99507

−2977

−177

5

15

3

−1

33

255

3183

100313

From the above data and using the above corollary we conclude that there is at least one root of in each of the intervals , and since changes sign in these intervals [i.e., for example ]. To locate a more accurate value of the root, we have to apply either the method of bisection, secant method or Newton’s method from numerical analysis. ■

CAS EXAMPLE 7

Prove that there is a root of the given equation in the specified interval by using the Intermediate Value Theorem.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-Exm7.html

Solution.

var('x')

f(x)= 2*x^3 - 5*x +10*cos(2*x)

plot(f, 0, 3)

bool ( f(1) < 0 ) # True

bool ( f(3) > 0 ) # True

Figure 6

Thus by the Intermediate Value Theorem, there exists at least one number such that . ■

2.2 EXERCISES (Continuity)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-2-2-Sol.html

http://youtu.be/azrkT1RP4-c http://youtu.be/hj8d-j_DGf4

CAS 1. Given continuous functions and with and

, find .

http://matrix.skku.ac.kr/cal-lab/cal-2-2-1.html

Solution. Since and are continuous functions.

var('g')

solve(3^2-4*3*g==5, g )

Answer : g(4)=1/3

2. Given continuous functions and with and , find .

Solution. .

Since is continuous, .

CAS 3.Show that the function is discontinuous at .

http://matrix.skku.ac.kr/cal-lab/cal-2-2-3.html

Solution. We may draw with the following Sage command. It shows the function diverges (+infinity) at . This shows is discontinuous at .

p1=plot(abs(ln((x-2)^2)), x, -3, 2, color='blue');

p2=plot(abs(ln((x-2)^2)), x, 2, 3, color='red');

show(p1+p2, ymax=5, ymin=-1)

If we use the following Sage command, we may get limit(abs(ln((x-2)^2))) at directly. It shows is discontinuous at .

limit(abs(ln((x-2)^2)), x=2)

Answer : +Infinity

4-7. Determine the points of discontinuity of . At which of these numbers is continuous from the right, from the left or neither? Sketch the graph of .

http://matrix.skku.ac.kr/cal-lab/cal-2-2-4.html

Solution. We may draw with the following Sage command. It shows the function is discontinuous at but continuous from the left.

plot(piecewise([[(-2*pi, 0), sin(x)], [(0, pi), cos(x)], [(pi, 2*pi), -1]]))

Solution. We see that exists for all except . Notice that the right and left limits are different .

Solution. We see that exists for all except and . Notice that the right and left limits are different at and .

8-10. For what values of the constant is the function continuous on ?

Solution.

Thus, for to be continuous on

CAS 9.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-9.html

Solution.

var('c')

solve(16-c^2==4*c+20, c)

Answer : c=-2

10.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-10.html

Solution. Suppose that function is continuous on . Find such that Find such that .

is continuous at (But is not continuos at ). The solution is .

var('c')

solve(10/(c-2)==2*c+4, c)

Answer : c=-3, c=3

11-13. Show that the following functions that has a removable discontinuity at . Also find a function that agrees with for and is continuous on ℝ.

11.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-11.html

Solution.

var('x,y')

p1=plot((x^3-2*x-4)/(x-2), x, -10, 10, color='blue');

show(p1, aspect_ratio=1, ymax=10, ymin=-10)

does not exist. Hence has removable discontinuity at . Then we can change the expression to remove the discontinuity. Let . It agrees with for and is continuous on ℝ.

12.

Solution. for . The discontinuity is removable and agrees with for and is continuous on ℝ.

13.

Solution. for .

The discontinuity is removable and agrees with for and is continuous on ℝ.

14.Let . Is removable discontinuity?

Solution. Since

, is not a removable discontinuity.

15. If , show that there is a number such that .

Solution. is continuous on the interval , and . Since , there is a number in such that by the Intermediate Value Theorem.

16. Prove using Intermediate Value Theorem that there is a positive number such that .

Solution. Let . We know that is continuous on the interval , and . Since , there is a number in such that by the Intermediate Value Theorem.

17-22. Prove that there is a root of the given equation in the specified interval by using the Intermediate Value Theorem.

CAS 17.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-17.html

Solution. Let . Then . Since is a real-valued continuous function on the interval , there is a root such that by the Intermediate Value Theorem.

var('x'); f(x)=x^4+x-3

plot(f, 1, 2)

f(1)

Answer : -1

f(2)

Answer : 15

18. ,

http://matrix.skku.ac.kr/cal-lab/cal-2-2-18.html

Solution. Let . Then . Since is a real-valued continuous function on the interval , there is a root such that by the Intermediate Value Theorem.

var('x')

f(x)=sqrt(6*x-x^2)-1

P=plot(f, 0, 6)

show(P, aspect_ratio=1)

bool(f(0)<0)

Answer : True

bool(f(2)>0)

Answer : True

CAS 19.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-19.html

Solution. Let . Then . Since is a real-valued continuous function on the interval , there is a root such that by the Intermediate Value Theorem.

var('x')

f(x)=x^(1/3)-1+x

plot(f, 0, 1)

f(0)

Answer : -1

f(1)

Answer : 1

CAS 20. ,

http://matrix.skku.ac.kr/cal-lab/cal-2-2-20.html

Solution.

var('x')

f(x)=exp(x^3)-x^6

plot(f, -1, 1)

bool(f(-1)<0)

Answer : True

bool(f(1)>0)

Answer : True

21.

Solution. is continuous on the interval , and . Since , there is a number in such that by the Intermediate Value Theorem. Thus, there is root of the equation in the interval .

22.

Solution. is continuous on the interval , and . Since , there is a number in such that by the Intermediate Value Theorem. Thus, there is root of the equation in the interval .

23-26. Show that each of the following equations has at least one real root.

23.

http://matrix.skku.ac.kr/cal-lab/cal-2-2-23.html

Solution. We may draw both and in one graph to find intersections. It shows the function has two real roots in and .

p1=plot(exp(x), x, -2, 4, color='blue')

p2=plot(4*sin(x), x, -2, 4, color='red')

show(p1+p2, ymax=5, ymin=0)

The following Sage commands give the value of the intersection in each interval, or .

find_root(exp(x)==4*sin(x), 0, 1)

Answer: 1.3649584337330951

24.

Solution. Let . Then and . By the Intermediate Value Theorem, there is a number in such that . This implies that .

25.

Solution. Let . Then and . By the Intermediate Value Theorem, there is a number in such that . This implies that .

26.

Solution. Let . Then and , and is continuous in . By the Intermediate Value Theorem, there is a number in such that

. This implies that .

p1=plot(exp(x), x, -2, 4, color='blue')

p2=plot(4*sin(x), x, -2, 4, color='red')

show(p1+p2, ymax=5, ymin=0)

The following Sage commands give the value of the intersection in each interval, or .

find_root(exp(x)==4*sin(x), 0, 1)

Answer : 0.37055809596982464

27-28.Find the values of for which is continuous.

27.

Solution. Suppose is continuous at in ℝ, then .

If is rational, then but or can be 1. If is irrational, then but or can be 0, this is a contradiction. Thus is discontinuous over ℝ.

28.

Solution. is continuous only at since . But there is no nonzero in ℝ, such that .

Calculus

About this book : http://matrix.skku.ac.kr/Cal-Book/

http://matrix.skku.ac.kr/sglee/ and http://matrix.skku.ac.kr/cal-book/

*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).

	1.9	1.99	1.999	1.9999	1.99999	1.999999
	3.9	3.99	3.999	3.9999	3.99999	3.999999

	2.1	2.01	2.001	2.0001	2.00001	2.000001
	4.1	4.01	4.001	4.0001	4.00001	4.000001

		−10	−5	−3	−2	−1	0	1	2	3	5	10
		−99507	−2977	−177	5	15	3	−1	33	255	3183	100313