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.

Examining the graph above, we need to look at three things:

1) What is the limit of the function as  approaches zero from the left?

2) What is the limit of the function as  approaches zero from the right?

3) What is the function value as  and is it the same as the result from statement one and two?

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

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.

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

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.

This particular question is looking for a one sided limit specifically a right handed limit. This is depicted by the plus sign in the exponent of the zero. Since we are looking for a right hand limit we want to look at the function values for x values that are slightly larger than zero.

Examining the graph, we can observe that  as  approaches  from the right.

This particular question is asking us to find a one sided limit. More specifically, it is asking us to find the left handed limit which means we want to look at the function values for x values that are slightly less than zero. This is depicted by the negative sign that is in the exponent on the zero.

Examining the graph, we can observe that  as  approaches  from the left.

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.

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=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 n into the remaining equation.

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-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 n into the remaining equation.

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve: