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 .

Step 1: To find vertical asymptotes of a rational function, we set the denominator to . We do this because a fraction is undefined with the denominator is , therefore an asymptote has been calculated..



Factor:



Set each parentheses equal to  and find :




Step 2: We said in Step 1 that any values that we got after setting the denominator to zero would be the asymptotes..

So, the asymptotes of  are .

It can also be written as .

Vertical asymptotes exist where the function divides by zero.  Therefore, it would be tempting to say there is a vertical asymptote at   However, we can factor the numerator of this function (difference of squares), which cancels out the bottom factor.  Therefore, there is no vertical asymptotes of this function.  At , the function has a hole.

Vertical asymptotes occur when the denominator of a function equals zero.  Therefore, we need to solve to find when that happens. 

Remember, whenever we have quadratic solutions, there are always positive and negative roots.

Horizontal asymptotes occur when the degree of the polynomial matches between the numerator and denominator or when the degree of the denominator polynomial is less than the numerator degree.  Here, we have the former case.  The horizontal asymptote in this case is the ratio of the coefficients in front of the leading polynomial.

Since we are looking for a horizontal asymptote, this refers to a  value.

Step 1: To find the horizontal asymptote, imagine what happens to the graph when x is very big. 



Step 2: After taking the limit of the function as x approaches infinity, we can see that the graph will eventually give me 

The line,  is also known as the x-axis.