Calculus Concept Collection - Chapter 2

Introduction to Limits

Answers

1. See the vocabulary section above.

2. a. means ( )f x has a vertical asymptote at x=-6, i.e., it gets larger

without bound as x approaches -6;

b. has a horizontal asymptote at y=-6, i.e., ( )f x is bounded by y=-

6 as x gets larger.

3. means that ( )f x has a horizontal asymptote at y=200, i.e., ( )f x

approaches 200 as x get larger.

4. means that as x approaches 175 from the left or the right, ( )f x

approaches 175.

Evaluate the following limits, if they exist. If a limit does not exist, explain why.

5.

22 (

73 )

3 7lim lim lim

88(1 )

3t t t

tt t

tt

t

tt

6. lim 3 3t

7. 2 4 4

2

1lim( ) lim ( 1)t t

t t tt

8. 2lim 2x

x x x

9. 2 2

2

5 7 9 5 7 9( )

( 1)( 3)2 3

g g g gh g

g gg g

, with vertical asymptotes at g=-1 and g=3;

2

2

2

2

7 95 )

li

(

(

m ( )

1

lim 52 3

)x x

gg g

h g

gg g

, means a horizontal asymptote at .

Given: perform the following:

10. 2

2

6 ( 3)( 2) ( 3)( )

( 4)( 2) ( 4)2 8

x x x x xf x

x x xx x

has a hole at (-2,

5

6), and a vertical asymptote

at x=4.

( 3)lim ( ) lim 1

( 4)x x

xf x

x

, means a horizontal asymptote at y=1.

Since 4

( 3)lim

( 4)x

x

x

and

4

( 3)lim

( 4)x

x

x

,

4

( 3)lim

( 4)x

x

x

is not defined.

11. 2

2

6( )

2 8

x xf x

x x

has a root at x=3, y-intercept at (0,

3

4 ), and hole at (-2,

5

6).

12. For , 1

lim 0nt t

13. For , 1

limnt t

14. For , 0

1 1lim l m 1

1i

t tt

15. If the degree of G is less than the degree of H, ( )

lim 0( )x

G x

H x .

16. If the degree of G is greater than the degree of H, ( )

lim( )x

G x

H x is either or .

17. If the degree of G is the same as the degree of H, ( )

lim( )x

G x

H x is a constant.

18. a. The amount of salt (in grams) after t minutes is 30 25t . The volume of water (in liters)

after t minutes is 8000 25t . Therefore the concentration, ( )C t , of salt after t minutes is

30 25

( )8000 25

tC t

t

.

b. 30 25 25 30

lim ( ) lim lim 303208000 25

25 ( 1)t t t

t tC t

tt

t

g/L. This makes sense since the longer

the pool fills, the more the concentration of salt in the pool looks like the additive.

One-sided Limits

Answers

1. 3

5limx

2. 2

lim 3x

3. 1

8limx

and 1

2limx

.

4. 1

im 2lx

5. 2

2limx

and 5

5limx

6. 2

2 2

2 8 ( 4)( 2)lim lim 6

2 2x x

x x x x

x x

7. 3

lim ( ) 1x

g x

8. 2

0 0

4lim li ( 4) 4m

x x

xx x

x

9. 1

lim ( ) 5x

g x

10. 2

1 1

4 3 (4 3)( 1)lim lim 7

1 1x x

x x x x

x x

11. 3

lim ( ) 4x

f x

12. 2

0 0

4lim lim ( 4) 4

x x

x xx

x

13. 2

lim ( ) 12x

h x

14. 2

2 2

4 7 2 (4 1)( 2)lim lim 9

2 2x x

x x x x

x x

15. 2

lim ( ) 7x

g x

16. 3

lim ( ) 9x

g x

17. 2

5 5

3 13 10 (3 2)( 5)lim lim 17

5 5x x

x x x x

x x

18. 2

lim ( ) 3x

f x

Properties of Limits

Answers

1. Use the addition rule: 2 2

2 2 2 2lim (3 4 9) lim 3 lim 4 lim 9 5x x x x

x x x x

2. Use the product rule. As 0x :

lim5 1x , cos(lim ) 1x , so then lim(5 cos( )) 1 1 1x x .

3. Use the addition rule. As 1x :

lim5 5x , cos(lim ) 0x , so then lim(5 cos( )) 5 0 5x x .

4. No. The limit will not exist if ( )g x does not have a limit.

5.

22

3

3

3

lim 7 127 12 42lim 7

3 lim 3 6

x

x

x

x xx x

x x

6. 3 3 2 3 2 3

32 2

lim 4 3 lim( 4 3) 27 3x x

x x x x

7. 3

3 2 2

2 2 2lim 4 (2 1) lim [ 4] ( lim [2 1]) 8 25 200x x x

x x x x

8. 2

7

7 12lim

7x

x x

x

does not exist because the denominator is 0 at 7x .

9. Yes. Since both functions are continuous, their limits are equal to their value. The product of

two continuous functions is continuous, so its value is equal to its limit as well. By the

product rule, the limit of the product is the product of the limits.

10. By definition, ( ( )) ( ) ( ) ( )g f x f x f x f x . Now, we know that the limit of ( )f x as 0x

is five. Apply the product rule and the addition rule to find that: 2lim ( ( )) 5 5 25 5 30g f x

11. Anything raised to the zero power is one, and cosine of zero is one. Therefore use the limit

rules to find that the limit is equal to:

0 0 0 25 2 2 5 4 9

cos(2

0)5e

12. e

. Plugging pi into the formula and using the limit rules gives us:

9 5 4 9 5 4

0(cos(sin(

3sin( )2 (cos(s))) in( )))

10 9 1 13

1 0 9e e

(cos( )0 )e e

13. By the limit rules, the limit of ( )f x must be a root of the polynomial 3 2( ) 1p x x x x .

We have to factor the polynomial: 3 2 21 (( 1)) ( 1)p x xx x xx The only root is at

one, so the limit of ( )f x as 0x is 1.

14. You cannot take the limit as x approaches zero because this would be dividing by zero.

15. Use the product rule. 5( ) lim ( ) ( ) ( ) ( ) ( ) lim (l ) lim ( )lim ( )limi ( ) li (m m )x f x f x f x f x f x f x f x f x f x f xf

5(lim ( ))f x . This method is not enough to prove the product rule because it will not prove

the case where the function is raised to a non-integer power.

Limits Involving Infinity

Answers

1) The degree of the numerator is greater than the degree of the denominator, so the function

will grow without bound. Since the denominator is x - 7, the function cannot include x = 7,

because the function cannot be defined where the denominator = 0. Logically, as x gets huge,

the -7 matters less and less, and we end up with just x2.

2) Similar to the last problem, the numerator is of greater degree than the denominator, so the

function does not approach a limit. The denominator is x - 3, so the graph cannot include 3.

3) As , or in other words, "as x gets huge", the value of x3 grows even faster, either

positive or negative, so there is no limit.

4) As x grows huge, x2 grows much faster than the rest of the expression, therefore, we can

approximate the end behavior of with

5) is a 3rd degree equation, so it will turn twice, since it is not a rational

function, there are no concerns about numerator or denominator. The function will have no

limits, and will grow without bound in both the positive and negative directions. If you use a

graphing calculator to graph the function, you will see that can be used to approximate

it.

6)

7)

8)

9)

10)

11)

12)

13) Zero at ; vertical asymptotes at ; as

14) Zero at ; no vertical asymptotes; as ; as

15) Zero at ; no vertical asymptotes but there is a discontinuity at ;

as ; as

Limits of Polynomial and Rational Functions

Answers

Solve the following rational function limits.

1. 2

1 1

12 1[ 3( 1

2lim lim

4 4)] 6

x x

x

xx

2. 2 2

3 11

3 9lim lim oes 2 4 18( 3)( 2)

11 6

x x

x D Not Exi tx

sx x x

(One-sided limits do not agree):

2

3 11

3 9lim2 4x

x

x

and

2

3 11

3 9lim2 4x

x

x

3. 57 57

56 56

5 3 13

2 3 6lim lim 0.17356 57 12 1

1

8x x

x

x

x x

4. 2

2 2

2 5 2lim lim

2(2 1) 3

x x

x xx

x

5. 2

3 3

4 4

4 5 6lim lim

4 3

11( 2)

4x x

xx

x

x

6. 4 4

3 3

2 3 5lim lim6 24 30(2 3)

6

( 4)

25

x x

x DNEx x x

x

4

3 3

2 3 5lim6 24x

x

x

and 4

3 3

2 3 5lim6 24x

x

x

.

7. 3 3

2 2

4 3 5

2 2 2lim lim2 3 2(2 3)(

8

1)x x

x

x DNEx x x

x

3

2

4 3 5

2 2 2lim2 3

x

x

x

x

and 3

2

4 3 5

2 2 2lim2 3

x

x

x

x

.

8. 2

4 4

8( 4

16lim li ) 0m

4x x

x x

xx

9. 10 10

39 39

3 3 7

3 4 6lim lim 0.05239 10

1

6(3 4)x x x

x

x

x

10. 2

4 4

3 7 20lim lim

4(3 5) 17

x x

xx

x

x

11. 2

4 4

4 14 8lim lim

4(4 2) 18

x x

x xx

x

12. 18 18

13 13

4 1 5

3 5 7lim lim 0.01613 1

1

8 7(3 5)x x

x

x

x x

13. 2 2

3 3

4 2lim lim 13 6 3

3

6

6

x x

x x

x x

14. 23 3

4 4

3 11

4 3 2lim lim4 3

46 27

2(4 3)x x

x

x

xx

x

15. 2

1 1

4 4

8 2 1lim lim

4

3( 1)

212

x x

xx x

x

16. 2

1 1

4 4

16 16 3lim lim

4 1(4 3) 2

x x

x x

xx

17. 2

0 0

3lim lim( 3) 3x x

x xx

x

Limits Involving Radical Functions

Answers

1. 3

lim 3x

x

2. 8

lim 7 1x

x

3. 4

2 1lim

4 4x

x

x

4. 1

3 2 1lim

1 4x

x

x

5. 0

lim 21 1x

x

x

6. 2 5lim 5

2xx x x

7. 6 2

3

3 1 1lim

44 3x

x x

x

8. 0

5 5 5lim

10x

x

x

9. 2

3lim( 4 ) 21x

x x

10. 2 3/2

1lim ( 2 10) 27x

x x

11. 4

5 3lim

64

1

x

x

x

12. 3 2

1lim( 2 3 7) 2 3x

x x

13. 3 2

3lim( 2 10) 2x

x

14. 7

lim2

5 35

3x

x

x

15. 2

0

4im 0

7 9lx

x

x

Limits Involving Trigonometric Functions

Answers

1. /6

1lim sin( )

2xx

2. /4

cotlim ( ) 1x

x

3. /

2

3seclim ( ) 4

xx

4. /3

1 3lim sin( ) co

2[ s( )]

xx x

5. /2

lim sec( ) tan( ][ ) x

x x Does Not Exist

6. lim cos 1

sin

xDoes Not Exist

x

x

7. The constant can be pulled out of the limit:

0 0

sin( ) 1 sin( ) 1 1lim lim 1

3 3 3 3x x

x x

x x .

8. 2 can be factored from the numerator and pulled out of the limit:

0 0 0

2cos( ) 2 2(cos( ) 1) cos( ) 1lim lim 2lim 2 0 0x x x

x x x

x x x

.

9. To evaluate this limit, we need a 3 in the denominator to match the one inside the sine

function. To introduce one, we must multiply both the numerator and denominator by 3:

0 0

sin(3 ) 3sin(3 )lim lim

3x x

x x

x x

And remove the 3 in the numerator from the limit.

0 0

3sin(3 ) sin(3 )lim 3lim 3 1 3.

3 3x x

x x

x x

10. Let1

yx

. Then as , 0x y . The limit becomes

0 0

1 1 sin( )lim sin( ) lim sin( ) lim 1x y y

yy

x yx

y

11. 2

21

cos( 1) 1lim 0

1x

x

x

(Use

2 1u x substitution and 0u in the limit)

12. Since the numerator approaches 1 as the denominator approaches zero, this expression will

not approach a finite value. Specifically,

0

cos( ) 1lim

0x

x

x

0

cos( ) 1lim

0x

x

x

So the limit does not exist.

13. 0 0

2lim(2 cot 2 ) cos 2 1

sl

ini

2m

x x

xx x x

x

.

14. 23/2 3/2 3/2

sin(2 3) sin(2 3) sin(2 3)lim lim lim

2 3 (2 3)( 1) 2 3

1 2 2(1) ( )

51 5x x x

x x x

x x x x x x

15. 0 0

tan sin 1lim lim 1 11

cosx x

x x

x x x

Limits of Composite Functions

Answers:

1. 2( ) 3f x x g h , where ( ) 3g x x and 2( )h x x .

2. 2( ) 4f x x x g h , where ( )g x x and 2( ) 4h x x x .

3. 2 3/2( ) ( 2 10)f x x x g h , where 3/2( )g x x and 2( ) 2 10h x x x .

4. 2( ) sin( 3)f x x g h , where ( ) sing x x and 2( ) 3h x x .

5. sin( ) 2 xf x g h , where ( ) 2xg x and ( ) sinh x x .

6. 2 2

2 2lim( 3) lim( ) 3 7x x

x x

7. 2 2

3 3lim( 4 ) lim( 4 ) 21x x

x x x x

8. 2 3/2 2 3/2 3/2

1 1lim ( 2 10) [ lim ( 2 10)] [9] 27x x

x x x x

9. 3 2 3 2

1 1lim( 2 3 7) lim(2 3 7) 2 3x x

x x x x

10. 2 2

/2 /2lim sin( ) sin[ lim ( )] 1

x xx x

11. lim (sin )

sin 1/2

/2lim 2 2 2 2

xx x

x

12. 1 1

2lim sin(2 ) sin[ lim (2 )] sin[ ]

2 2 4 2

x x

x x

13. 3 2 2 3

33 3

lim( 2 10) lim(2 10) 8 2x x

x x

14.

2lim (2 )22 21

1lim

tt x

xe e e

15. 3 2

2

1 1 1

1 ( 1)( 1)lim lim lim ( 1) 3

1 1x x x

x x x xx x

x x

Continuity of a Function

Answers

1. While graph of the function appears to be continuous everywhere, a check of the table

values indicates that the function is not continuous at .

2. While the function appears to be continuous for all , a check of the table values

indicates that the function is not continuous at .

3. Test for continuity by evaluating the limit from either side, then evaluating the function.

3 3 3lim lim ( 3)) l| 3 | ( im 3 (3) 3 0x x x

xx x

3 3lim lim 033 | 3 3|x x

xx

(3) | 3 3 | | 0 | 0f

Because 3 3lim lim( ) ( ) (3)x x

f x f x f

, the function is continuous at the point.

4. Test for continuity by evaluating the limit from either side, then evaluating the function.

22

6 1

(2 5

6 1( )

4 2 )0 25

xf x

x x

x

x

5/2 5/2lim lim( ) ( )

x xf x f x

2(

5)f does not exist. Therefore the function is not continuous.

5. Test for continuity by evaluating the limit from either side, then evaluating the function. 2

3 2 2

3 13 10 (3 2)( 5)

4 2 15 ( 5)( 3)

)(

x x x x

x x x x x xf x

5 5

17lim lim

33( ) ( )

x xf x f x

(5)f does not exist.

The function is not continuous at the point.

6. 2k

7.

(a) 3 3

lim lim 1 1 ( 3) 4 2( )x x

f xx

3 3

4 4lim lim 6 ( 3) 6 4 6 2

3( )

3x xf xx

( 3)f does not exist, but since the limits agree with one another, 3x is a

removable discontinuity.

(b) 3 3

4 4lim lim 6 (3) 6 4 6 1) 0

3(

3x xxf x

2 2

3 3lim lim 3 9( )x x

f x x

(3) 10f Even though we can evaluate the function at 3x , the limits do not agree with

each other so 3x is a jump discontinuity.

8. For both parts (a) and (b), it will be useful to have the function in factored form. 3 2 2 2

2

3 27 9 (3 1) 9(3 1) ( 9)(3 1) ( 3)( 3)(3 1)

(3 1)( 1) (3 1)( 1) (3 1)( 1)3 2 1( )

x x x x x x x x x x x

x x x x x xx xf x

(a) At 1x the function’s numerator approaches a finite value while the denominator

approaches 0. This is a pole type discontinuity.

(b) Observe that ( 3)( 3)(3 1)

(3 1)( 1(

))

x xf x

x

x x

is equal to ( 3)( 3)

( 1)

x x

x

everywhere that

they both exist; however, ( )f x does not exist at1

3x . Therefore

1 1

3 3

( 3)( 3) 182lim li( ) m

( 1) 3x x

xf

x

xx

.

Since the limit exists, though the function does not, the discontinuity is removable.

9. The function will be discontinuous only where the denominator equals zero. We find these

points by setting each term in the product equal to zero and find 0x and 1

4x .

2 1 0x has no solution. The fact that 4 1x also appears in the denominator does not

mean there is no discontinuity due to this term; it only means that it is a removable

singularity instead of an asymptote. Since the function is defined everywhere except 0x

and 1

4x , The intervals on which x is continuous are (

1, )

4 , (

1,0)

4 and (0, ) .

10. Because the sum of continuous functions is continuous, we look at the continuity of each

term separately. 3 has no discontinuities. 2

3x is discontinuous only at 3. Therefore the

intervals of continuity of ( )f x are ( ,3) and (3, ) .

11. cot( )x is undefined when tan( ) 0x . 2cot ( )x is the same. tan( ) 0x at ,2 ,3, .0 . . any

integer multiple of . Therefore ( )f x is undefined at these points, and the intervals of

continuity of ( )f x are , ),( ,0),(0, ),( ,2 ),(2 ,3 ),... .. 2 .( There are infinitely

many.

12. False. ( )f x is continuous everywhere its denominator is nonzero. But, the denominator is

zero at 1

2x which is in the interval [0,1]. ( )f x is not continuous on this interval.

13. False. The denominator of ( )f x is zero at / 3 which is in the interval [ / 4 , / 2 ]. Since

the function is undefined at / 3 it is not continuous there.

14. 2( ) 4 2f x x x is undefined, therefore not continuous, in the open interval

(2 2) (2 2)x because the integrand of the function is <0 in that interval. It is

continuous everywhere else.

15. 2( ) 4 5f x x x is defined everywhere and therefore continuous everywhere.

Properties of Continuous Functions

Answers

1. f(-3)=-5 and f(-2)=3, so there is at least one root in the interval

2. f(9)<0 and f(10)>0, so there is at least one root in the interval.

3. f(-3)>0 and f(0)<0, so there is at least one root in the interval.

4. f(-0.5)>0 and f(0)<0, so there is at least one root in the interval.

5. f(-2.75)>0 and f(0)<0, so there is at least one root in the interval.

6. True. First observe that this function is continuous everywhere, so it is continuous on the

interval (0, ) . Next, test the endpoints of the interval. c(0) sin(0 o) s(0) 0 1 1f ,

co( s) (si ( )) 0 1n 1f . So the IVT guarantees a root on the interval (0, ) .

7. False. The IVT can only guarantee that certain intervals will contain roots; it says nothing

about which intervals will not contain roots.

8. False. Although we can confirm that (0) 0f and )( 0f , The function is discontinuous

where the denominator, cos( ) 0x . This occurs at 2

x

, which is on the interval (0, ) .

Therefore, the IVT does not apply to this problem. Furthermore, we can guarantee that this

function has no roots anywhere on the real line since its numerator is never equal to zero.

9. We can confirm that this function is continuous everywhere, so it is continuous on all of the

above intervals. Next we check the values of ( )f x at the endpoints. (0) 0 03

, )(2

f f

so

the IVT makes no guarantee about the interval (0,2

)3

. ( , () 0 ) 0

3 3f f

so, again, the

IVT says nothing about the interval in option (b).

)(2

03

f

, however, and we already know (0) 0f . So there is a root on the

interval (2

)03

,

. Since (2

)03

,

is contained in ( ), , we know there is also a

root on this interval without examining its endpoints. The correct answers are (c)

and (d).

10. Answers will vary. First confirm that this function is continuous on any given interval. Next,

note that 0xe everywhere, so ( ) 0f x for any 0x . Thus 0x is a good right endpoint

for our interval. It remains to find a left endpoint with ( ) 0f x . 1x is a fine choice; here

1( 1)2.7

12

1 11 1 0

ef e . So one interval is (-1,0). Most negative numbers work

for the left endpoint, and any positive number for the right endpoint, as well.

11. Intervals that include the function’s roots at 5, 2,2x are correct.

12. False. The function is not continuous on the interval [-3, 3].

13. True. The function is continuous on the interval (1, 3).

14. True. The function is continuous on the interval (-3, 0).

15. False. The function is not continuous on the interval [-3, 0].