When finding horizontal asymptotes, there are 3 conditions / rules to follow.

1) if the leading terms of the numerator and denominator are of the same degree, then the HA is equal to the ratio of the coefficients of the leading terms

2) if the leading term of the denominator is of a higher degree than the leading term of the numerator, then the HA is 

3) if the leading term of the numerator is of a higher degree than the leading term of the denominator, then there is no horizontal asymptote (or it is undefined)

This question falls into condition 2, and therefore the horizontal asymptote is .

When finding horizontal asymptotes, there are 3 conditions / rules to follow.

1) if the leading terms of the numerator and denominator are of the same degree, then the HA is equal to the ratio of the coefficients of the leading terms

2) if the leading term of the denominator is of a higher degree than the leading term of the numerator, then the HA is 

3) if the leading term of the numerator is of a higher degree than the leading term of the denominator, then there is no horizontal asymptote (or it is undefined)

This question falls into condition 3, and therefore there is no horizontal asymptote / the asymptote is undefined.

The denominator is not valid and does not exist at .  There is an asymptote at .  However, the question asked for the limit when the graph is approaching  from the right side.  

Following the curve of the graph of , the range will approach negative infinity when we follow the graph from the right end of .

Therefore, the answer is: 

The easiest way to determine this limit is to reduce tangent to its component parts.

As x approaches  from the left, the numerator approaches  while the denominator approaches , meaning that the limit is positive infinity. 

Therefore,

The denominator equals zero only where . 

 If  is negative. Then

.

Since is negative, then is positive, and by plugging in , we get 

.

Also, if , then 

.

Therefore  must be a positive non-zero integer. Let us analyze two separate cases: one where is an even positive integer, and the other where is an odd positive integer. Suppose is even. Then we can write for some positive integer . Then our limit becomes

.

Now lets look at the graph of

shown below.

As we can see, the left and right limits are approaching . This is backed up by the following calculations:

.

Since both left and righthanded limits agree, then we can conclude that .

Therefore, if is even, we can conclude that  Therefore the limit is for positive nonzero even integers. Now suppose is a positive odd integer. Then we can write in the form of , where is any non-negative integer.

Thus we have the following:Now let us examine the graph of , which is shown below.

Here, it appears that the left-hand limit and the right-hand limit will not coincide. This is supported by the following calculations:

Since the left and right limits do not exist, then the full limit does not exist.

Thus we have 

Therefore the limit does not exist for positive odd integers.

We can conclude that must be an even non-zero positive integer.

Step 1: Describe what kind of function we have here..

We have a 5th power function...

Step 2: Describe when we have asymptotes:

We have asymptotes if the function we are analyzing is a rational fraction..

 is not a rational function...there is no asymptotes to this function.

Evaluating the numerator, it gets closer and closer to -1 as x gets closer to -3. The denominator approaches 0. -1/0 suggests either negative or positive infinity as a solution.

Since we are approaching -3 from the positive side, the denominator will be positive. So, -1 divided by a very small positive number gives us .

Evaluating the numerator, we get 2-1=1. By evaluating the denominator, regardless of whether we approach from the negative side or the positive side, we get a very small number that is squared. Since the number is squared, it will always be positive.

So, 1 divided a small positive number approaches .