A vertical asymptote occurs at  when 

or .

In our case, since we have a quotient of functions, we need only check for values of  that make the denominator , but don't also make the numerator 



This equals  when  is an integer multiple of .

Hence the vertical lines  are vertical asymptotes.

However we must exclude the case , because this will also cause the numerator to be , thus creating a "hole" instead of an asymptote.

Hence our answer is

.

We note that for all , we have .

Hence,

By inverting the above inequality and multiplying by x. We get the following:

.

We know that,

and by the Squeeze Theorem,

we have:

We first need to see that the function sin(x) has infinitely many roots.

We can express these roots in the following form:

, wkere k is an integer.

The function has the roots as asymptotes.

Therefore this function's vertical asymptotes are expresses by , where k is an integer. Since the integers are infinitely many, the vertical asymptotes are infinitely many.

To find the above limit, we need to note the following.

We have for all n positive integers:

.

(We can verify this formula by the long division)

Now we need to note that:

, where .

We have then:

and we have

.

Since,

we obtain the following:

We need to notice that the function f is defined for all real numbers.

We need to also remark that for all reals:

implies that

this gives again:

and therefore,

.

This function can't be 0.

Assume for a moment that

, this implies that but this cannot happen since we are dealing with real numbers.

Therefore the above function can never be 0 and this means that it does not have a vertical asymptote. This is what we needed to show.

For this problem we first need to expand the denominator.

We can expand the denominator since  is a difference of squares.

From here we can cancel the  quantity from the numerator and denominator.

The resulting function is as follows:

Plugging in 2 we get our limit.

We will use the following to prove this result.

Assuming that

. We will use this result:

we have

Therefore

this shows the limit is 1.

We will use the following identity to establish this result.

We have

and we note that :

Therefore by multiplying the above equivalency for 1 we get the following:

 and we know that

We can rewrite our equation using identities. 

This gives :

 and

now taking the limit as x goes to 3, we obtian

We note first that we can write:

Therefore our expression becomes in this case:

Noting now that:

for all .

Therefore , we have: 

and evaluating now for x=1, we obtain

We first note that the polynomial is defined for all real numbers.

We know that for any real number x different from 0, we have :

.

Now we need to see that for any integer n we have:

. Adding in this case,

we have

and therefore , this implies by definiton of q(x) that:

.

We also have .

This means that  .

Therefore q(x) can never be 0 and this means that it does not have an asymptote.