Answer. Note that we have an indeterminate form . You may want to try Hopital rule and see how difficult
the calculations become except if you use the substitution .
Here we will use Taylor Polynomials to find this limit. We have

This gives

Clearly the limit is

Problem 2. Find the limit

Answer. Note that we have an indeterminate form . Since we a have a square root in the function, we will use
the conjugate forms to get rid of it. We have

Since

(note that |x| = -x because and therefore x
< 0), hence

It is now obvious that

Problem 3. Find

Answer. Note that we have an indeterminate form . In
order to find the limit, we will use the formula

We can assume that . In this case, we have

Since

we have to concentrate on the term . We have

Since

we get

Therefore we have

Problem 4. Determine the convergence or divergence of

(a)

Answer. The bad points are , , -1.
At , we have

when . Recall that

when if and only if

Using the limit test and the p-test, we know that

are both convergent. Thereofre

are both convergent.
At -1, we have

when . Again using the limit test and the
p-test, we know that

is divergent. Therefore the improper integral

is divergent.

(b)

Answer. The bad points are 0 and .
At , we have

when . Using the limit test and the
p-test, we know that

is divergent. Therefore

is divergent. So the improper integral

is divergent. We suggest to consider the other bad points too though
there is no need for that. But you never know, your previous conclusion may
be wrong!!!!
At 0, we have

when . Using the limit test and the
p-test, we know that

is divergent. Therefore

is divergent. So again we get the divergence of the hole improper integral.

(c)

Answer. The bad points are 1 and .
At 1, we have

when . Using the limit test and the
p-test, we know that

is convergent.

At , we have

when . And since

when and the improper integral

is divergent, then using the limit test and the
p-test, we have

is divergent. We deduce then that the improper integral