Formal Definitons of Limits

(3/1908)Section 1.5

Formal definitions of limits

Overview: The definitions of the various types of limits in previous sections involve phrases such as

arbitrarily close, sufficiently close, arbitrarily large, and sufficiently large. For instance, according

to Definition 2 of Section 1.1, limxa

f(x) = L with a number L if f(x) is arbitrarily close to L for all x 6= asufficiently close to a. These qualitative formulations of the definitions are all that are needed in most of

calculus. More quantitative formulations of the definitions are required, however, in dealing with difficult

examples and in proofs. In these formal definitions, phrases such as arbitrarily close and sufficiently

close are made precise by using inequalities. In this section we discuss the formal definition of two-sided

finite limits limxa

f(x) = L. Formal definitions of limx

f(x) = L and limx

f(x) = are presented inthe context of infinite sequences in Section 11.1. Other definitions are similar and are studied in advanced

courses.

Topics:

The -definition of a finite two-sided limit Using approximate inequalities Two short proofs Using graphs with finite limits

The -definition of a finite two-sided limitTo convert Definition 2 in Section 1.1 of lim

xaf(x) = L into a formal definition, we replace the phrase

f(x) is arbitrarily close to L with |f(x) L| < for an arbitrarily small positive number , andwe replace the phrase for all x 6= a sufficiently close to a with for all x with 0 < |x a| < for asufficiently small positive number . We obtain the following:

Definition 1 (Finite two-sided limits) Suppose that y = f(x) is defined on open intervals (b, a) and

(a, c) on both sides of a. Then limxa

f(x) = L with a number L if and only if for every positive , there is

a positive such that

|f(x) L| < for all x with 0 < |x a| < . (1)

and are the Greek letters epsilon and delta.

51

p. 52 (3/1908) Section 1.5, Formal definitions of limits

Statement (1) means that the portions of the graph y = f(x) for a < x < a and for a < x < a+are between the lines y = L+ and y = L as is illustrated in Figure 1.

x

yy = f(x)

a+

y = L+

y = L

a

L

a

|f(x) L| < for

0 < |x a| < FIGURE 1

To see the geometric meaning of this Definition 1, imagine that the positive parameter in Figure 1

decreases toward zero, so that the lines y = L and y = L+ approach the line y = L. If the definitionis satisfied, then for each , there is a such that the portion of the graph for a < x < x0 on the leftof a and the portion for a < x < a+ on the right of a lie between those lines. Since can be arbitrarily

small, this causes the value f(x) of the function to approach L as x approaches a from both sides.

Notice that if a has been found to satisfy Definition 1 for a particular , then any smaller positive

would also work for that and the same would serve for any larger .

We can often show that Definition 1 is satisfied by giving a rule that defines a suitable for every

positive , as in the following example.

Example 1 Figure 2 shows the graph of the function y = F (x) defined by

F (x) = 4x for x 6= 2.Use Definition 1 to prove that lim

x2F (x) = 8.

x1 2 3

y

4

8

12

y = F (x)

x2

y

8

y = F (x)

8 +

8

2 2

limx2

F (x) = 8|F (x) 8| <

for 0 < |x 2| <

FIGURE 2 FIGURE 3

Section 1.5, Formal definitions of limits p. 53 (3/1908)

Solution We need to show that for each positive number there is a positive number such that

|F (x) 8| < for all x with 0 < |x 2| < (2)

(Figure 3). We do this in three steps. We begin with preliminary calculations. Next,we use these calculations to define a rule that provides a suitable for each . Then weverify that the rule works.

(Preliminary calculations) For x 6= 2, F (x) equals 4x and

|F (x) 8| < |4x 8| < 4|x 2| < |x 2| < 1

4.

(3)

(The rule) Given > 0, set = 14 so that the last expression in (3) is the same as

the last expression in (2).

(Verification) With this definition of , statement (2) is valid because if x satisfies0 < |x 2| < , then it satisfies the last inequality in (3) and therefore |F (x) 8| < .

Example 2 Show, using Definition 1, that limx0

x3 = 0.

Solution Since |x3 0| = |x3| and |x 0| = |x|, we need a rule defining a positive for eachpositive (Figure 4) such that

|x3| < for all x with 0 < |x| < . (4)

(Preliminary calculations) We solve the first inequality in (4) for |x| by taking cuberoots of both sides:

|x3| < |x| < 1/3 (5)

(The rule) Given > 0, define = 1/3 so that the last expressions in (4) and (5) arethe same.

(Verification) Statement (4) holds with this definition of because if 0 < |x| < ,then |x| < 1/3 and |x|3 < by (5).

x

yy = x3

y =

y =

|x3| < for

0 < |x| < FIGURE 4


Example 3 Use Definition 1 to prove that the statement limx0

x3 = 2 is false.

Solution We need to show that there is a positive such that there is no positive with theproperty that |x3 2| < for all x with 0 < |x| < .

Since the limit is actually 0, we can use any < 2. We will use = 1. There isno positive such that |x3 2| < 1 for all x with 0 < |x| < since 0 < x3 < 1 andhence |x3 2| > 1 for 0 < x < 1.

Using approximate inequalitiesIn Examples 1 and 2 we found formulas for to satisfy the definition of lim

xaf(x) = L by solving the

inequality |f(x)L| < exactly. In many cases, this inequality is relatively difficult to solve and we use,instead, the following procedure in which this inequality is replaced by one that is easier to deal with.

(Preliminary calculations) Use factoring or other techniques to write f(x) L in the formf(x) L = g(x)(x a)

with another function y = g(x). Then pick a specific positive number 1 (1 = 1 will often work) suchthat g(x) is bounded for 0 < |x a| < 1 and pick a positive constant M such that |g(x)| M for0 < |x a| < 1. Then

|f(x) L| M |x a| for 0 < |x a| < 1. (6)(The rule) The positive number must be defined so that

|f(x) L| < for all x with 0 < |x a| < . (7)Accordingly given > 0, let be the smaller of 1 and /M .

(Verification) Statement (7) is valid with this definition of because if 0 < |x a| < , then|f(x) L| M |x a| < since 1 , and then |f(x) L| < since /M .Example 4 Show that by Definition 1, lim

x3x2 = 9.

Solution We need a rule that gives a positive for every positive such that

|x2 9| < for 0 < |x 3| < . (8)(Preliminary calculations) We start by factoring |x2 9| in the form |x+ 3||x 3|.Since |x + 3| is bounded in any finite interval, we set 1 = 1. Then 2 < x < 4 for|x 3| < 1 and, consequently, |x+ 3| < 7 for |x 3| < 1. Therefore,

|x2 9| = |x+ 3||x 3| 7|x 3| for |x 3| < 1. (9)(The rule) Inequality (9) shows that we can have |x2 9| < , as in (8), by requiring|x 3| < 1

7 so that 7|x 3| < and by requiring |x 3| < 1, so that (8) holds. Given

> 0, we define to be the smaller of 1 and 17.

(Verification) Statement (8) holds with this definition of because if

0 < |x 3| < , then |x 3| < 1 so that |x2 9| 7|x 3| by (9) and then

|x2 9| 7|x 3| < 7 7(1

7)= .

The procedure in Example 4 gives, for each > 0, a that satisfies Definition 1, even though itis generally not the largest possible such for that . For example, the inequality |x2 9| < 1 holds forpositive x if 8 < x2 < 9 or equivalently if

8 < x 0, let be the minimum of 12and 1

2.

(Verification) Statement (10) is valid with this choice of because if 0 < |x 1| < ,then 1

2 x 3

2since 1

2and then by (11)

1x 1 2|x 1| < 2 2( 12 ) = .

since 12.


Two short proofsThe next two examples illustrate how Definition 1 is used in proofs.

Example 6 Use Definition 1 to prove that if limx8

g(x) = 10, then limx8

2g(x) = 20.

Solution We need a rule that gives a positive for every positive such that

|2g(x) 20| < for all x with 0 < |x 8| < . (12)

Since the inequality on the left of (12) can be written 2|g(x) 10| < and this isequivalent to |g(x) 10| < 1

2, we apply Definition 1 to lim

x8g(x) = 10 with 1

2 in

place of . According to that definition, there is a > 0 such that |g(x) 10| < 12 for

0 < |x 8| < , and this shows that (12) holds. Example 7 Suppose that lim

xx0f(x) = L and lim

xx0g(x) =M with numbers L and M . Prove, using

Definition 1, that limxx0

[f(x)+g(x)] = L+M . (This is the first statement in Theorem 1

of Section 1.1.)

Solution We need to show that for every > 0 there is a > 0 such that

|[f(x) + g(x)] (L+M)| < for all x 0 < |x x0| < . (13)

(Preliminary calculations) Because f(x) tends to L and g(x) tends to M , |f(x)L|and |g(x) M | are arbitrarily small for all x 6= x0 sufficiently close to x0. We use thetriangle inequality

|A+ B| |A|+ |B| (14)

which is valid for any numbers A and B, to write

|[f(x) + g(x)] (L+M)| = |[f(x) L] + [g(x)M ]| |f(x) L|+ |g(x)M | . (15)

(The rule) Let be an arbitrary positive number. The number on the right of (15) is

less than if |f(x) L| < 12 and |g(x) M | < 1

2. From Definition 1 with replaced

by 12 and applied to f and g, there are positive numbers 1 and 2 such that

|f(x) L| < 12 for 0 < |x x0| < 1

|g(x)M | < 12 for 0 < |x x0| < 2.

(16)

We defiine to be the smaller of 1 and 2.

(Verification) With this definition of , statement (13) holds because if0 < |x x0| < , then inequalities (16) hold and by (15),

|[f(x) + g(x)] (L+M)| |f(x) L|+ |g(x)M | < 12+ 1

2 = .

Inequality (14) is valid because |A+B| = |A| + |B| if A and B are both 0 or are both 0, and |A+B| < |A|+ |B|if one of A and B is positive and the other is negative. This result is called the triangle inequality because it is the statement

PR PQ +QR that the length of one side of a triangle PQR is no greater than the sum of the lengths of the other two sides in

the special case where P = (A, 0), Q = (0, 0), and R = (B, 0) and PQR is a line segment.


Using graphs with finite limitsSuitable values of in Definition 1 for specific s can sometimes be found from graphs, as in the nextexample.

Example 8 According to Definition 1 applied to the limit limx1

(x+ 3

x) = 2 with = 0.25, there

is a positive such that

|(x+ 3x) 2| < 0.25 for all x with 0 < |x 1| < . (17)

Use the graph of y =x+ 3

x in Figure 6 to find the approximate value of the largest

that could be used in (17).

x1 2

y

1

2

y =x+ 3

x

FIGURE 6

Solution Since 2 0.25 = 1.75 and 2 + 0.25 = 2.25, (17) is the same as the statement

1.75 0, a > 0 suchthat |(5x 6) 14| < for 0 < |x 4| < . Find the largest possible such for = 0.01.

Verify Definition 1 or 2 for the limits in Exercises 3 through 5.

3.O limx5

3x 5 = 0

4.A limx3

(6x+ 2) = 20

5. limx1

(x2 + 2x) = 3

6.O According to Definition 1 applied to the limit limx1

(1

x+

1

x2

)= 2 with = 0.5, there is a

positive such that(1

x+

1

x2

) 2

< 0.5 for all x with 0 < |1 x| < . ()(a) Use the graph of y =

1

x+

1

x2in Figure 10 to find the approximate value of the largest that

could be used in (). C(b) Check the result of part (a) with a graphing calculator or computer.

x1 2

y

1

2

3

y =1

x+

1

x2

FIGURE 10

Verify Definition 1 for the limits in Exercises 7 through 14.

7.O limx5

x2 = 25

8.A limx0

6

x+ 2= 3

9. limx4

(3x+ 6) = 18

10. limx2

(x2 + 3) = 7

11.A limx5

15

x= 3

12.O limx2

8

x2= 2

13.A limx0

30x

x+ 7= 0

14. limx1

3x

x+ 2= 1


In Exercises 15 through 18, (a) use the graph to find the approximate value of the largest that can

be used in Definition 1 with the given value of . C(b) Check your result with a graphing calculator orcomputer.

15.O limx1

(4 4x2 + 2x3) = 2 with = 0.4; Figure 11

x1

y

1

2

3

5

y = 4 4x2 + 2x3

FIGURE 11

16.A limx1

(1

x+ x2

)= 2 with = 0.5; Figure 12

x1 2

y

1

2

3

y =1

x+ x2

FIGURE 12

17.A limx4

(x2 +x) = 18 with = 4; Figure 13

x1 2 3 4

y

10

20

30y = x2 +

x

FIGURE 13


18. limx2

80

x+ 2= 20 with = 2; Figure 14

x1 2 3

y

20

10

y =80

x+ 2

FIGURE 14

EXPLORATION:

19.O (a) The width w of a cube is measured to be 2 centimeters and this value is used to calculate

the volume V = w3 = 8 (cubic centimeters) of the cube. How accurately must the width bemeasured in order that the resulting error in the calculated volume be 0.1 cubic centimeters?(b) What does this calculation have to do with the definition of a limit?

20.A (a) A hunk of clay weighs exactly 100 grams, and its volume is measured to be 50 cubiccentimeters. This value is used to calculate the density = 100/V = 2 (grams per cubiccentimeter) of the clay. How accurately must the volume be measured in order that the resultingerror in the calculated density be 0.05 grams per cubic centimeter? (b) What does thiscalculation have to do with the definition of a limit?

21. (a) The area A of a square is measured to be 9 square meters and this value is used to calculate

its width w =A =

9 = 3. How accurately must the area be measured in order that the

resulting error in the calculated width be 0.1 meters? (b) What does this calculation have todo with the definition of a limit?

C In Exercises 22 and 23, use a graphing calculator or computer to find the approximate value of the thelargest that could be used in Definition 1 with the given value of in the definition of the limit

C 22.A limx1

(4 + 4x ln x) = 4 with = 0.5

C 23. limx1

[x+ 2 sin( 12pix)] = 3 with = 0.5

24. Does = 0.2 work with = 0.2 in Definition 1 applied to limx1

x2 = 1?

Use Definition 1 to prove that the statements in Exercises 25 through 28 are false.

25.O limx3

x2 = 9.2

26.O limx0.01+

1

x=

27.A limx1

x2 = 0

28. limx0

(x2 + 5) = 0

29. Define G(x) = x for rational numbers x and G(x) = 0 for irrational x. Use Definition 1 to showthat (a) lim

x0G(x) = 0 and (b) the two-sided finite limit lim

xx0G(x) does not exist for x0 6= 0.

30. Suppose that limxx0

f(x) = L and limxx0

g(x) = M with numbers L and M . Use Definition 1 to

show that (aO) limxx0

[f(x)g(x)] = LM and (b) limxx0

f(x)

g(x)=

L

Mif M 6= 0.

(End of Section 1.5)

Formal Definitons of Limits

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