If we were to look at a plot of the function, the limit would be obvious. Drawing the graph would be difficult without a graphing device. We can overcome this by first defining a new variable and assessing how it behaves as the original variable approaches  from the right.  

Let,

Therefore,  as 

We notice that the function in terms of  has a graph that is much easier to draw from recollection since we readily know how the natural logarithm behaves. In the plot of the natural log, the y-axis is a horizontal asympote, and the function becomes infinite in the negative direction as  becomes arbitrarily close to 0 from the right. Hence  . 

The graph above is more familiar. This is not the graph of the original function , which is in terms of  The plot above represents the function defined in terms of the variable we defined, . 

As  gets closer to 3 (but remains larger than 3), then  gets closer to 0 (but remains a small positive number).

The numerator, , gets closer to 6.

So,  is an arbitrarily large positive number. 

Thus, we conclude that the limit is .

As  gets closer to 3 (but remains smaller than 3), then  gets closer to 0 (but remains a small negative number).

The numerator, , gets closer to 6. So,  is an arbitrarily large negative number. 

Thus, we conclude that the limit is .

Recall the graph of a natural log function.

Loosely speaking, .

Graphing the function , is as follows and it is seen that as the function approaches three from the right hand side the function values approach negative infinity. 

Since the first derivative of  approaches  from both the left and right side of , the function should decrease on both sides of .

The only graph that is decreasing throughout is .

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.