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=1; 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. We see that we can no longer factor this to make the denominator not equal 0; hence this limit DNE because the denominator is zero.

$\displaystyle \\ \lim_{x \to 1}\frac{3x^2+3x}{x^2-1}\\ \\=\lim_{x \to 1}\frac{3x(x+1)}{(x-1)(x+1)}\\ \\=\lim_{x \to 1}\frac{3x}{x-1}\\ \\=\frac{3}{0}\Rightarrow DNE$

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: $\displaystyle -\infty$  from the left and $\displaystyle \infty$ 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, 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, we can observe that $\displaystyle \lim_{x\rightarrow 5}f(x)$ does not exist, as   is not continuous at $\displaystyle x=5$. 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 $\displaystyle c=5$, we can see that condition #1 is not satisfied because the graph has a vertical asymptote instead of only one value for $\displaystyle f(5)$ and is therefore an infinite discontinuity at $\displaystyle (5,f(5))$.

We can also see that condition #2 is not satisfied because $\displaystyle \lim_{x\rightarrow 5}f(x)$ approaches two different limits: $\displaystyle -\infty$  from the left and $\displaystyle \infty$ from the right.

Based on the above, condition #3 is also not satisfied because $\displaystyle \lim_{x\rightarrow 5}f(x)$ is not equal to the multiple values of .

Thus, $\displaystyle \lim_{x\rightarrow 5}f(x)$ does not exist.

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

1) What is the limit of the function as $\displaystyle x$ approaches zero from the left?

2) What is the limit of the function as $\displaystyle x$ approaches zero from the right?

3) What is the function value as $\displaystyle x=0$ and is it the same as the result from statement one and two?

Thus, we can observe that $\displaystyle \lim_{x\rightarrow 0}f(x)=-\infty$  as $\displaystyle x$ approaches $\displaystyle 0$ from the left and from the right.

First we need to determine if this is a left or right sided limit. Since there is a plus sign in the exponent on the zero that means it is a right sided limit. This means we want to look at the function values for x values that are slightly larger than zero.

Therefore examining the graph, we can observe that $\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=\infty$  as $\displaystyle x$ approaches $\displaystyle 0$ 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 $\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=\infty$ as $\displaystyle x$ approaches $\displaystyle 0$ from the left.

First we need to determine what the question is asking. Since there is a plus sign in the exponent on the zero this indicates that we are looking for a right sided limit. This means we want to look at the function values for values slightly larger than zero.

Examining the graph, we can observe that $\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=\infty$ as $\displaystyle x$ approaches $\displaystyle 0$ 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 $\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=\infty$ as $\displaystyle x$ approaches $\displaystyle 0$ 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 positive infinity.

Thus, we can observe that $\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=\infty$ as $\displaystyle x$ approaches $\displaystyle 0$ from the right.