Calculus I - Spring 2010

Midterm Exam, March 3, 2010

In the following problems you are required to show all your work and provide the necessary explana-

tions everywhere to get full credit.

1. Find the domain of the function f(x) =x + 1

x2 − 3.

Solution: This function does not exist at points where the denominator is 0, that is at x = −√

3and x =

√3. Therefore the domain is

{

x | x 6= ±√

3}

or(

−∞,−√

3)

∪(

−√

3,√

3)

∪(√

3,∞)

2. Let f(x) =x − 4

|4 − x| . Find the value of each limit, if it exists. (If it does not exist, enter

NONE.)

(a) limx→4−

f(x) = −1

Solution: We have

limx→4−

x − 4

|4 − x| = limx→4−

x − 4

4 − x= lim

x→4−

x − 4

−(x − 4)= −1

(b) limx→4+

f(x) = 1

Solution: We have

limx→4+

x − 4

|4 − x| = limx→4+

x − 4

−(4 − x)= lim

x→4+

x − 4

x − 4= 1

(c) limx→4

f(x) does not exist

Solution: Since

limx→4−

x − 4

|4 − x| 6= limx→4+

x − 4

|4 − x|it follows that lim

x→4f(x) does not exist.

1

3. Use the given graph of f(x) to find the value of each limit, if it exists. (If it does not exist,enter NONE.)

(a) limx→−4−

f(x) = 2

(b) limx→−4+

f(x) = 2

(c) limx→−4

f(x) = 2

(d) limx→1−

f(x) = 4

(e) limx→1+

f(x) = −2

(f) limx→1

f(x) NONE, since limx→1−

f(x) 6= limx→1+

f(x)

(g) limx→6−

f(x) = 5

(h) limx→6+

f(x) = 5

(i) limx→6

f(x) = 5

2

4. For each of the following, decide if the limit exists. If it does, find the limit. If it does not,decide also if the limit is ∞, −∞, or neither.

(a) limx→

√5

x2 − 5

x −√

5= 2

√5

Solution: We have

limx→

√5

x2 − 5

x −√

5= lim

x→√

5

(x −√

5)(x +√

5)

x −√

5= lim

x→√

5(x +

√5) =

√5 +

√5 = 2

√5

(b) limx→1

x2 − 4x + 3

(x − 1)2does not exist and neither ∞ nor −∞

Solution: We have

limx→1

x2 − 4x + 3

(x − 1)2= lim

x→1

(x − 1)(x − 3)

(x − 1)2= lim

x→1

x − 3

x − 1

Note that

limx→1−

x − 3

x − 1= ∞ and lim

x→1+

x − 3

x − 1= −∞

Since

limx→1−

x − 3

x − 16= lim

x→1+

x − 3

x − 1

it follows that limx→1

x2 − 4x + 3

(x − 1)2does not exist and neither ∞ nor −∞.

(c) limx→0

sin(8x)

tan(2x)= 4

Solution: We have

limx→0

sin(8x)

tan(2x)= lim

x→0

sin(8x)sin(2x)cos(2x)

= limx→0

(

cos(2x) · sin(8x)

sin(2x)

)

= limx→0

cos(2x)︸ ︷︷ ︸

=1

· limx→0

sin(8x)

sin(2x)=

= limx→0

sin(8x)

sin(2x)= lim

x→0

sin(8x)8x

· 8xsin(2x)

2x· 2x

=

[

limu→0

sin u

u= 1

]

= limx→0

1 · 8x1 · 2x = lim

x→0

8

2= 4

(d) limx→∞

5x2 − x + 1

3x2 + 2=

5

3

Solution: We have

limx→∞

5x2 − x + 1

3x2 + 2= lim

x→∞

5x2−x+1x2

3x2+2x2

= limx→∞

5x2

x2 − xx2 + 1

x2

3x2

x2 + 2x2

= limx→∞

5 − 1x

+ 1x2

3 + 2x2

=5 − 0 + 0

3 + 0=

5

3

3

(e) limx→∞

(√

x2 + 2x −√

x2 + x) =1

2

Solution: We have

limx→∞

(√

x2 + 2x −√

x2 + x) = limx→∞

√x2 + 2x −

√x2 + x

1=

= limx→∞

(√

x2 + 2x −√

x2 + x)(√

x2 + 2x +√

x2 + x)√x2 + 2x +

√x2 + x

= limx→∞

(x2 + 2x) − (x2 + x)√x2 + 2x +

√x2 + x

=

= limx→∞

x√x2 + 2x +

√x2 + x

= limx→∞

xx√

x2+2xx

+√

x2+xx

= limx→∞

1√

x2+2x√x2

+√

x2+x√x2

=

= limx→∞

1√

x2+2xx2 +

√x2+x

x2

= limx→∞

1√

x2

x2 + 2xx2 +

√x2

x2 + xx2

= limx→∞

1√

1 + 2x

+√

1 + 1x

=

=1√

1 + 0 +√

1 + 0=

1

2

(f) limx→1

1

(1 − x)2= ∞ (does not exist)

Solution: We have

limx→1−

1

(1 − x)2= lim

x→1+

1

(1 − x)2= ∞

therefore

limx→1

1

(1 − x)2= ∞ (does not exist)

5. Use the Squeeze Theorem to find the limit limx→∞

sin x

x.

Solution: We have

−1 ≤ sin x ≤ 1 =⇒ −1

x≤ sin x

x≤ 1

x

Since

limx→∞

(

−1

x

)

= limx→∞

1

x= 0

it follows that

limx→∞

sin x

x= 0

by the Squeeze Theorem.

REMARK: Do not confuse limx→∞

sin x

x= 0 and lim

x→0

sin x

x= 1.

4

6. Find the horizontal and vertical asymptotes of the function f(x) =x2 + 4x − 5

x2 + x − 2.

Solution:

(a) We have

limx→−2−

x2 + 4x − 5

x2 + x − 2= lim

x→−2−

(x − 1)(x + 5)

(x − 1)(x + 2)= lim

x→−2−

x + 5

x + 2= −∞

Therefore x = −2 is the vertical asymptote.

REMARK: Note that x = 1 is NOT a vertical asymptote, since

limx→1−

x2 + 4x − 5

x2 + x − 2= lim

x→1−

(x − 1)(x + 5)

(x − 1)(x + 2)= lim

x→1−

x + 5

x + 26= ±∞

Similarly,

limx→1+

x2 + 4x − 5

x2 + x − 2= lim

x→1+

(x − 1)(x + 5)

(x − 1)(x + 2)= lim

x→1+

x + 5

x + 26= ±∞

(b) We have

limx→±∞

x2 + 4x − 5

x2 + x − 2= lim

x→±∞

x2+4x−5x2

x2+x−2x2

= limx→±∞

x2

x2 + 4xx2 − 5

x2

x2

x2 + xx2 − 2

x2

= limx→±∞

1 + 4x− 5

x2

1 + 1x− 2

x2

=1 + 0 − 0

1 + 0 − 0= 1

Therefore y = 1 is the horizontal asymptote.

7. Let f(x) = sin x.

(a) Find an equation of the tangent line to f(x) at the point (0, 0).

Solution: We have

f ′(x) = (sin x)′ = cos x =⇒ f ′(0) = cos 0 = 1

therefore slope m = 1. It follows that an equation of the tangent line to f(x) at the point (0, 0)is

y − 0 = 1 · (x − 0) =⇒ y = x

(b) Find an equation of the normal line to f(x) at the point (0, 0).

Solution: Since

mnormal = − 1

mtangent= −1

1= −1

it follows that an equation of the normal line to f(x) at the point (0, 0) is

y − 0 = −1 · (x − 0) =⇒ y = −x

5

8. Find the derivative of the function f(x) =2√x

using the definition of derivative.

Solution: We have

f ′(x) = limh→0

f(x + h) − f(x)

h= lim

h→0

2√x+h

− 2√x

h= lim

h→0

2√

x−2√

x+h√x√

x+h

h= lim

h→0

2√

x − 2√

x + h

h√

x√

x + h=

= limh→0

(2√

x − 2√

x + h)(2√

x + 2√

x + h)

h√

x√

x + h(2√

x + 2√

x + h)= lim

h→0

4x − 4(x + h)

h√

x√

x + h(2√

x + 2√

x + h)=

= limh→0

−4h

h√

x√

x + h(2√

x + 2√

x + h)= lim

h→0

−4√x√

x + h(2√

x + 2√

x + h)=

=−4√

x√

x + 0(2√

x + 2√

x + 0)=

−4√x√

x(2√

x + 2√

x)=

−4

x · 4√x= − 1

x√

x= −x−3/2

9. Let f(x) = sin(sin x). Find f ′(x) and f ′′(x).

Solution: We have

f ′(x) = [sin(sin x)]′ = cos(sin x) · (sin x)′ = cos(sin x) · cos x

andf ′′(x) = [cos(sin x) · cos x]′ = [cos(sin x)]′ · cos x + cos(sin x) · (cos x)′ =

= − sin(sin x) · (sin x)′ · cos x + cos(sin x) · (− sin x) =

= − sin(sin x) · cos x · cos x − cos(sin x) · sin x =

= − sin(sin x) · cos2 x − cos(sin x) · sin x

10. Use limits to explain why the function

f(x) =

x2 − x + 2 if x ≤ 2

1

x + 2if x > 2

is discontinuous at x = 2.

Solution: We have

limx→2−

f(x) = limx→2−

(x2 − x + 2) = 22 − 2 + 2 = 4 and limx→2+

f(x) = limx→2+

1

x + 2=

1

2 + 2=

1

4

Sincelim

x→2−f(x) 6= lim

x→2+f(x)

it follows that f(x) is discontinuous at x = 2.

6

11. Differentiate the following functions:

(a) f(x) = 7√

2x

Solution: We have

f ′(x) = (7√

2x)′ = 7(√

2x)′ = 7((2x)1/2)′ = 7 · 1

2(2x)1/2−1 · (2x)′ =

= 7 · 1

2(2x)−1/2 · 2 = 7 · (2x)−1/2 =

7√2x

(b) f(x) = x(1 − x)8

Solution: We have

f ′(x) = [x(1 − x)8]′ = x′(1 − x)8 + x[(1 − x)8]′ = (1 − x)8 + x · 8(1 − x)7 · (1 − x)′ =

= (1 − x)8 + 8x(1 − x)7 · (−1) = (1 − x)8 − 8x(1 − x)7 = (1 − x)7(1 − 9x)

(c) f(x) =1 − 3

√x

5x3

Solution: We have

f ′(x) =

(1 − 3

√x

5x3

)′

=

(1 − 3x1/2

5x3

)′

=

(1

5x3− 3x1/2

5x3

)′

=

(1

5x−3 − 3

5x1/2−3

)′

=

=

(1

5x−3 − 3

5x−5/2

)′

=1

5

(x−3

)′ − 3

5

(x−5/2

)′=

1

5· (−3)x−3−1 − 3

5·(

−5

2

)

x−5/2−1 =

= −3

5x−4 +

3

2x−7/2

(d) f(x) =2 cos x − 4 sin x

x + 1

Solution: We have

f ′(x) =

(2 cos x − 4 sin x

x + 1

)′

=(2 cos x − 4 sin x)′ · (x + 1) − (2 cos x − 4 sin x) · (x + 1)′

(x + 1)2=

=(−2 sin x − 4 cosx) · (x + 1) − 2 cos x + 4 sinx

(x + 1)2

(e) f(x) = sec2(1 +√

5 − x)

Solution: We have

f ′(x) = [sec2(1 +√

5 − x)]′ = 2 sec(1 +√

5 − x) · (sec(1 +√

5 − x))′ =

= 2 sec(1 +√

5 − x) · sec(1 +√

5 − x) tan(1 +√

5 − x) · (1 +√

5 − x)′ =

= 2 sec(1 +√

5 − x) · sec(1 +√

5 − x) tan(1 +√

5 − x) · 1

2(5 − x)−1/2 · (−1) =

= − sec2(1 +√

5 − x) · tan(1 +√

5 − x) · (5 − x)−1/2

7