A vertical asymptote occurs atwhen

or.

In our case, since we have a quotient of functions, we need only check for values ofthat make the denominator,但也别让分子



This equalswhenis an integer multiple of.

Hence the vertical linesare 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 functionhas 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 thatbut 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 sinceis a difference of squares.

From here we can cancel thequantity 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:

和评估x = 1,我们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.