Calculus Practice Makes Perfect Ch1

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1

LIMITS

Te undamental idea o the calculus is the concept o limit. Te exercises in Part Iare designed to improve your understanding and skills in working with this con-cept. Te symbolisms involved are useul contractions/abbreviations and recogniz-ing the “orm” o these is essential in successully producing required results. Beoreyou begin, i you need a review o unctions, see Appendix A: Basic unctions andtheir graphs.

· I ·



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3

Limit definition and intuitionA unction f x ( ) is said to have a limit A as x approaches c written lim ( )

x c f x A

l ,

provided the error between f x ( ) and A, written | ( ) | f x A , can be made less thanany preassigned positive number E whenever x is close to, but not equal to, c. Heu-ristically, “Te limit o f at the point c is A i the value o f gets near A when x is near c.”We will explore this defnition intuitively through the ollowing examples.

Compute the value o f x x ( ) or the ollowing values o x that are closeto, but not equal to 2 in value; and then make an observation about theresults.

a. x 2.07 f x ( ) .b. x 1.98 f x ( ) .c. x 2.0006 f x ( ) .

Observation: It appears that when x is close to 2 in value, then f (x ) is close to9 in value.

Compute the value o f x x

( )

or the ollowing values o x that are close to,

but not equal to 0 in value, and then make an intuitive observation about theresults.

a. x .01 f x ( ) b. x –.001 f x ( ) –c. x .001 f x ( )

Observation: It appears that when x is close to 0 in value, f (x ) is not close toany fxed number in value.

Using limit notation, you can represent your observation statements or theabove examples, respectively, as:

limx

x l

and limx x l

does not exist.

The limit concept ·1·



4 Limits

1·1

EXERCISE

Compute the value o (x) when x has the indicated values given in (a) and (b). For (c), make

an observation based on your results in (a) and (b).

1. x ( ) x

x

2

5

a. x 3.001

b. x

2.99

c. Observation? ________________________________________________________

2. x ( ) x

x

5

4

a. x 1.002

b. x .993

c. Observation? ________________________________________________________

3. x ( ) 32 x

x

a. x .001

b. x .001

c. Observation? ________________________________________________________

Properties of limitsBasic theorems that are designed to acilitate work with limits exist, and these theorems are the“bare bones” ideas you must master to successully deal with the limit concept. Succinctly, themost useul o these theorems are the ollowing:

I lim ( )x c

f x l

and lim ( )x c

g x l

both exist, then

1. Te limit o the sum (or dierence) is the sum (or dierence) o the limits.

lim[ ( ) ( )]x c f x g x l lim ( ) lim ( )x c x c f x g x l l

2. Te limit o the product is the product o the limits.

lim[ ( ) ( )]x c

f x g x l

lim ( ) lim ( )x c x c

f x g x l l

3. Te limit o a quotient is the quotient o the limits provided the denominator limit is not .

lim( )

( )x c

f x

g x l

lim ( )

lim ( )x c

x c

f x

g x l

l

4. I f x f x f x x c

n

x cn( ) , thenq

l llim ( ) lim ( ) or n

5. lim ( ) lim ( )x c x c

af x a f x l l

where a is a constant

6. lim[ ( )] lim ( )x c x c

n

f x f x n

l l §

©¶¸ or any positive integer n

7. limx c

x cl

8. limx c x cl

provided c w



The limit concept 5

You must guard against the error o writing or thinking that lim ( )x c

f x l

f c( ); that is, that you

determine the limit by substituting x c into the expression that defnes f (x ) and then evaluate.Recall that in the limit concept, x cannot assume the value o c. Te complete explanation re-quires the concept o continuity, which is discussed in Chapter 3.

PROBLEMS Evaluate the ollowing limits.

a. limx

x

x l

b. lim( )x

x x l

c. limx

x

x l

SOLUTIONS a. limlim

limx

x

x

x

x

x

x l

l

l

b. lim( ) lim lim limx x x x

x x x x x l l l l

llim

x x l

c. lim lim

( )( )

lim(x x x

x

x

x x

x x l l l

)

Notice that in this example, you cannot use the quotient theorem because the limit o thedenominator is zero; that is, lim( )

x x

l

. However, as shown, you can take an algebraic ap-

proach to determine the limit. First, you actor the numerator. Next, using the act that or all

x x x

x x w

,

( )( ), you can simpliy the raction and then evaluate the limit. Tis is a

useul approach that can be applied to a number o limit problems.

d. limx

x l

does not exist because 6x 12 0 when x is close to 1.

1·2

EXERCISE

Find the ollowing limits or indicate nonexistence.

1. lim x

x

x l

3

24

16. lim

x

x

x l

0

2

3

9 3

11

2. lim x

x

x l

2

29

27. lim

x

x x

x l

1

2

2

2 1

1

3.lim x x l 1

3

78.

lim x

x

x l

4 2

6 3

16

4. lim( ) x

x l

P

5 92

9. lim x

x l

2

34 11

5. lim x

x

x l

0

5 3

1110. lim

x

x

x l

6

8 3

6



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7

Zero denominator limitsSome o the most useul limits are those in which the denominator limit is 0, eventhough our previous limit theorems are not directly applicable in these cases.

Tese types o limits can exist only i there is some sort o cancellation com-ing rom the numerator. Te key is to seek common actors o the numerator anddenominator that will cancel.


a. limx

x

x l

b. lim( ) ( )

h

x xh h x

hl

c. limh

x h x

hl

SOLUTIONS a. lim lim( )( )

limx x x

x

x

x x x

x l l l

( )x x

b. lim

( ) ( )

limh h

x xh h x

h

xh

l l

h

h

llim( )h

x h x

c. lim lim( )( )

( )h h

x h x

h

x h x x h x

h x h x l l

l llim

( )

( )lim

( )h h

x h x

h x h x

h

h x h x

l

lim( )h x h x x

PROBLEM I f (x ) 6x 2 7, then fnd lim ( ) ( )h

f x h f x hl

.

SOLUTION lim( ) ( )

lim( ( ) ) (

h h

f x h f x

h

x h x

l l

))

h

llimh

x xh h x

h

l

lim( )h

x h x

Special limits ·2·



8 Limits

2·1

EXERCISE

Evaluate the ollowing limits.

1. lim x

x

x x l

3 2

3

12

2. lim( )

h

x h x

hl

0

2 2

3. lim x

x

x l

4

3

2

64

16

4. I x x ( ) 5 8, fnd lim( ) ( )

h

x h x

hl

0

.

5. lim x

x

x l

3 2

5 7

3

6. lim x

x

x l

25

5

25

7. I g x x ( ) 2 , fnd lim( ) ( )

x

g x g

x l

2

2

2.

8. lim x

x x

x l

0

22 4

9. limr

x r x

r l

0

10. lim x

x

x l

4

36

4

Infinite limits and limits involving infinityTe variable x is said to approach c c( ) i x increases (decreases) without bound. For example, x approaches c i x assumes the values 2, 3, 4, 5, 6, and so on, consecutively. Note, however, that y does not approach infnity i y assumes the values 2, –2, 4, –4, 6, –6, and so on, in the samemanner.

A unction becomes positively infnite as x approaches c i or every M 0, f (x ) M orevery x close to, but not equal to, c. Similarly, a unction becomes negatively infnite as x ap-proaches c i or every M 0, f (x ) M or every x close to, but not equal to, c.


a. lim ,x

a

x lc

where a is any constant

b. limx

x lc

c. limx

x

x lc

d. lim| |x x l



Special limits 9

SOLUTIONS a. limx

a

x lc or any constant a

b. limx

x lc

c

c. lim limx x

x

x x

x

lc lc

d. lim| |x x l

c

2·2

EXERCISE

Evaluate the ollowing limits.

1. lim( ) x

x lc

5 7 6. lim x

x

x x lc

2

5 62

2. lim x x lc

73

7. lim x

x x

x x lc

5 3

6 2

6 7

5 6 11

3. lim x

x lc

3 95 8. lim x

x x x

x x lc

7 6 3

3 7 5

4 2

3

4. lim x

x x x

x x lc

3 2

3

47 9

18 76 119. lim

x

x x

x lc

2 8 5

3 4

3

2

5. lim x x lc

8

410. lim

x x lc 5

42

Left-hand and right-hand limitsDirectional limits are necessary in many applications and we write lim ( )

x c f x

l to denote the limit

concept as x approaches c through values o x larger than c. Tis limit is called the right-handlimit o f at c ; and, similarly, lim ( )

x c f x

l is the notation or the le-hand limit o f at c .

Teorem: lim ( )x c

f x Ll

i and only i lim ( ) lim ( )x c x c

f x f x Ll l

. Tis theorem is a very useul tool

in evaluating certain limits and in determining whether a limit exists.


a. limx x l

b. limx x l

c. lim[ ]x

x l

Note: [x ] denotes the greatest integer unction (See Appendix A)



10 Limits

d. lim[ ]x

x l

e. lim[ ]x

x l

SOLUTIONS a. limx x l

c

b. limx x l

c

c. lim[ ]x

x l

d. lim[ ]x

x l

e. lim[ ]x

x l

does not exist because lim[ ]x

x l

and lim[ ]x

x l

, so the right- and

le-hand limits are not equal.

2·3

EXERCISE

Evaluate the ollowing limits i they exist. I a limit does not exist, show why.

1. lim [ ] x

x l

4

1 6. lim x

x

x l

3

29

3

2. lim x

x

x l

2

24

27. lim

x x l 4

7

4

3. lim x x l 8

4

98. lim

x

x x

x l

4

5 48

4

4. lim x x l 0 4 3 9. lim x

x

x l

4

216

4

5. lim[ ] x

x l

5

1 10. lim x

x

x l

4

216

4



11

Continuity

Definition of continuityA unction f is continuous at a point c i and only i

1. f c( ) is defned; and

2. lim ( ) ;x c

f x existsl

and

3. lim ( ) (lim ) ( ).x c x c

f x f x f cl l

I a unction ails to satisy any one o these conditions, then it is not con-tinuous at x c and is said to be discontinuous at x c.

Roughly speaking, a unction is continuous i its graph can be drawn with-out liing the pencil. Strictly speaking, this is not mathematically accurate, but itis an intuitive way o visualizing continuity.

Notice that when a unction is continuous at a point c, you have the situationwhereby the limit may be calculated by actually evaluating the unction at thepoint c. Recall that you were cautioned against determining limits this way in anearlier discussion; however, when a unction is known to be continuous at x c,then lim ( ) ( ).

x c f x f c

l

By its defnition, continuity is a point-wise property o a unction, but thisidea is extended by saying that a unction is continuous on an interval a x ba ai and only i f is continuous at each point in the interval. At the end points, theright- and le-hand limits apply, respectively, to get right and le continuity i these limits exist.

PROBLEMS Determine whether the ollowing unctions are eithercontinuous or discontinuous at the indicated point.

a. f x x ( ) at x 4

b. f x x ( ) at x 3

c. f x x ( ) at x 2

d. f x x

( )

at x 2

e. f x x

x ( )

at x 2

·3·



12 Limits

SOLUTIONS a. lim lim( ) (lim ) ;x x x

x x x l l l

thus, the

unction is continuous at 4.

b. lim( ) ((lim ) ) ;x x

x x l l

thus, the unction is continuous at 3.

c. lim( ) ( (lim ) ) ;x x

x x l l

thus, the unction is continuous at 2.

d. limx x l

does not exist; thus, the unction is discontinuous at 2.

e. f x x

x ( )

is discontinuous at 2 because the unction is not defned at 2.

However, the limit o f (x ) as x approaches 2 is 4, so the limit exists but

lim ( ) ( ).x

f x f l

w

I f ( ) is now defned to be 4 then the “new” unction f (x )

x

x x

x

w

ª«-

¬-

¹º-

»-is continuous at 2. Since the discontinuity at 2 can be “removed,”

then the original unction is said to have a removable discontinuity at 2.

3·1

EXERCISE

Show that the ollowing unctions are either continuous or discontinuous at the indicated

point.

1. x x ( ) 5 7 at x 1 6. g x x

x ( )

5

5at x 3

2. x x

x ( )

38

2at x 0 7. g x

x

x ( )

5

2at x 8

3. x

x

( )

4

2 3

at x 1 8. h x x x ( ) 5 72

at x 5

4. x x ( ) [ ] at x 3 9. x x

x ( )

6

2at x 6

5. g x x

x ( )

26

5at x 4 10. h x

x a x

x a( )

( )

26

3at x a

Properties of continuityTe arithmetic properties o continuity ollow immediately rom the limit properties in Chapter 1.

I f and g are continuous at x c, then the ollowing unctions are also continuous at c:1. Sum and dierence: f o g

2. Product: fg

3. Scalar multiple: af , or a a real number

4. Quotient: f g , provided g c( ) w



Continuity 1

Further, i g is continuous at c and f is continuous at g (c) then the composite unction f g C

defned by ( )( ) ( ( )) f g x f g x C is continuous at c. In limit notation, lim ( ( )) (lim ( ))x c x c

f g x f g x l l

f g c( ( )). Tis unction composition property is one o the most important results o continuity.

I a unction is continuous on the entire real line, the unction is everywhere continuous;that is to say, its graph has no holes, jumps, or gaps in it. Te ollowing types o unctions arecontinuous at every point in their domains:

Constant unctions: f (x ) k, where k is a constant

Power unctions: f x x n( ) , where n is a positive integer

Polynomial unctions: f x a x a x a x an

n

n

n( )

!

Rational unctions: f x p x

q x ( )

( )

( ), provided p x ( ) and q x ( ) are polynomials and q x ( ) w

Radical unctions: f x x x n( ) , , q n a positive integer

rigonometric unctions: f (x ) sinx and f (x ) cosx are everywhere continuous; f (x ) tan x , f (x ) cscx , f (x ) secx , and f (x ) cotx are continuous only wherever they exist.

Logarithm unctions: f x x ( ) ln and f x x b bb

( ) log , , w

Exponential unctions: f x ex ( ) and f x b b bx ( ) , , w

PROBLEM Discuss the continuity o the ollowing unction: g x x ( ) (sin ) at a real number c.SOLUTION 3x is continuous at c and sinx is continuous at all real numbers and so sin( )x is

continuous at c by the composition property. Finally, sin( )x is continuous at cby the constant multiple property o continuity.

3·2

EXERCISE

Discuss the continuity o the ollowing unctional expressions.

1. x x ( ) ( tan ) 5 3 at a real number c 6. G x x x

x

( )sin

11 8 9

2

on the real line

2. h x x x ( ) tan cos( ) 3 1 at c 4 7. V x x x ( ) sin cos on the real line

3. x x x

x x x x ( )

costan sin

5 23

34at c 5 8. T x x x ( ) sin cos 2 2

at c P

11

4. t x x x ( ) cos tan 5 3 at c P

29. x

x

x ( )

tan

sin at x 2P and at x 6P

5. H x x x ( ) sin 8 4 1325

or x 1 10. g x x x ( ) sin 15 10 at x 11

Intermediate Value Theorem (IVT)Te Intermediate Value Teorem states: I f is continuous on the closed interval [a, b] and i f a f b( ) ( )w , then or every number k between f (a) and f (b) there exists a value x

in the interval

[a, b] such that f x k( )

.Te Intermediate Value Teorem is a useul tool or showing the existence o zeros o a unc-

tion. I a continuous unction changes sign on an interval, then this theorem assures you that there



14 Limits

must be a point in the interval at which the unction takes on the value o 0. It must be noted,however, that the theorem is an existence theorem and does not locate the point at which the zerooccurs. Finding that point is another problem. Te ollowing example will illustrate using the IVto determine whether a zero exists and give some insight into fnding such a point (or points).

PROBLEM Is there a number in the interval [0, 3] such that f x x x ( ) ?

Tis question is equivalent to asking whether there is a number in [0, 3] suchthat f x x x ( ) .

SOLUTION Te unction is continuous on [0, 3], and you can see that f ( )

and f ( ) . Since f ( ) and f ( ) , by the IV, you know

there must be a number in [0, 3] such that f x x x ( ) ; that is, there

is a solution to the problem. In this case, a solution can be ound by solving

the quadratic equation, x x 0, to obtain the two roots:

o.

Approximating these two values gives 1.62 and 0.62, o which only 1.62 is in the

interval [0, 3]. Tus, there does exist a number, namely

, in the interval

[0, 3] such that f

¤

¦

¥³

µ

´ .

3·3

EXERCISE

For 1–5, use the IVT to determine whether the given unction has a zero in the given interval.

Explain your reasoning.

1. x x x x ( ) 4 3 2 54 3

on [2, 0] 4. g x x

x ( )

8

3 5on [10, 12]

2. g x x ( ) 92

on [2.5, 2] 5. x x x

x

( )

3 2

1

on [2, 2]

3. x x

( ) 3

4on [5, 0]

For 6–10, use the IVT to determine whether a zero exists in the given interval; and, i so, fnd the zero

(or zeros) in the interval.

6. h x x x ( ) 25 2 on [3, 4] 9. F x x ( ) cos( ) on [5, 8]

7. g x x

x ( )

33

4on [0, 6] 10. G x

x

x ( )

sin( )

cos( ) on

§

©¨

¶

¸·

P P

4 4,

8. h x x ( ) sin( ) on [1, 1]

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Calculus Practice Makes Perfect Ch1