1. Find the limits (4 points each)

**Solutions**
(a) The limit
represents an indeterminate

form . Therefore we can keep only the leading
terms in the

numerator and in the denominator of the fraction .

. Because x tends to positive infinity we
have

whence .

(b) The limit represents an indeterminate
form . As

always when we deal with such an indeterminate form we will first

find the limit of the natural logarithm of our expression . We have

, and the limit
is an indeterminate

form because the first factor tends to ∞ and
the second one -

to ln 1 = 0. To apply the L'Hospital's rule we have to write the last

limit as an indeterminate form or
. In our case it is easier to write

it in the form , namely,
. According to the

L'Hospital's rule we will now differentiate the numerator and the

denominator and look at the following limit (when differentiating the

numerator we combine the chain rule and the quotient rule).

Therefore .

(c) We have here an indeterminate form
. First we bring it to

the form ,
.

Next we use the L'Hospital's rule to obtain the following limit

It is still an indeterminate form
so we apply the
L'Hospital's rule

once again and get the limit

(d)

This is an indeterminate form
and we apply the
L'Hospital's rule.

2. Find the derivatives of y = f(x) with respect to x (4
points each)

**Solutions.**

(a) f(x) = cos (sin (x + 1)). We apply the chain rule to get

f'(x) = -sin (sin (x + 1)) cos (x + 1):

(b) . By the quotient
rule combined with the chain rule

we have

We simplify the last expression by multiplying both
numerator and

denominator by .

(c) . By the product
rule combined with the chain

rule we have

If we recall the formula

then we can also write our answer as

(d) 5y^{2} + sin y = x^{2}. We apply implicit differentiation to
get

Solving it for we get

3. Find the integrals (5 points each)

**Solutions**

(a)

If we notice that x is proportional to the derivative of
3x^{2} + 1 we can

use the following substitution, u = 3x^{2} +1. Then
, du = 6xdx,

and therefore . The integral becomes

Applying the power rule we get

(b)

Notice that sec^{2} x equals to the derivative of tan x and
therefore it is

convenient to use the substitution u = tan x. Then du = sec^{2} xdx and

our integral becomes

(c) We apply the power rule and the Newton - Leibnitz
formula to get

(d)

We use the substitution u = t^{3} + 1. Then du = 3t^{2}dt whence

6t^{2}dt = 2du. If t = -1 then u = 0, and if t = 0 then u = 1, we will

change the limits of integ ration accordingly .

4. Use a linear approximation to estimate the following
values (5 points

each)

**Solutions** In both cases we use the formula for linear
approximation

L(x) = f(a) + f'(a)(x - a)

where point a should satisfy two (informal) conditions

•
a should be close to x

•
the values of f (a) and f0(a) should be easy to compute.

(a) In this case we take whence
. We also take

a = 64 and x = 63. Then f(a) = 4, , and x - a
= -1. The

corresponding linear approximation to
is

(b) Remember that we do not use the degree measure in
calculus. In

our case we take

Then

Therefore the value of the linear approximation is