To evaluate the limit, we must determine whether the limit is right or left sided. Because 0 has a negative sign "exponent", we know we are approaching from values slightly less than 0, or from the left side. Now, evaluate the limit using the part of the piecewise function corresponding to values less than (or equal to) 0, and you get .

The limiting situation in this equation would be the denominator. Plug the value that  is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=5; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=-3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=-2; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

This limit DNE because the denominator is zero and we cannot factor to get anything else.

Examining the graph, we can observe that does not exist, as   is not continuous at . We can see this by checking the three conditions for which a function  is continuous at a point :

1. A value  exists in the domain of 

2. The limit of  exists as  approaches 

3. The limit of  at  is equal to 

Given , we can see that condition #1 is not satisfied because the graph has a vertical asymptote instead of only one value for  and is therefore an infinite discontinuity at .

We can also see that condition #2 is not satisfied because  approaches two different limits:  from the left and  from the right.

Based on the above, condition #3 is also not satisfied because  is not equal to the multiple values of .

Thus, does not exist.

To evaluate the limit, we must first pull out a factor consisting of the highest power term divided by itself (so we are unchanging the contents of the limit):

After the factor we pulled out cancels to 1, we can see that the numerator of the fraction goes to zero (as infinity is reached to the -1 power).

Therefore, the limit approaches .

Examining the graph, we want to find where the graph tends to as it approaches zero from the left hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the left, the function values of the graph tend towards positive infinity.

Thus, we can observe that  , as  approaches  from the left.

Examining the graph, we want to find where the graph tends to as it approaches zero from the right hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the right, the function values of the graph tend towards positve infinity.

Therefore, we can observe that  , as  approaches  from the right.