When we evaluate the limit, we find that as \(\displaystyle tan(x)\) approaches \(\displaystyle \pi\), we approach \(\displaystyle 0\). Because there is a negative sign in front of the function (we were given \(\displaystyle -tan(x))\), the answer is still \(\displaystyle 0\).

The fact that the limit is right sided doesn't change the outcome.

We first need to determine what the question is asking for. In this case because there is a plus sign in the exponent of zero we can interpret that as the limit of the function as x values approach zero from the right hand side. This means the values are going to be 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.

First we want to check 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 satisfied because the graph is approaching negative infinity at .

We can also see that condition #2 is satisfied because  approaches  from the right and from the left.

Based on the above, condition #3 is also satisfied because  is equal to .

Thus by examining the graph, 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.

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.

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

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.

\(\displaystyle \\ \lim_{x \to 2}\frac{x^2+5x-14}{x^2-5x+6}\\ \\=\lim_{x \to 2}\frac{(x-2)(x+7)}{(x-2)(x-3)}\\ \\=\lim_{x \to 2}\frac{x+7}{x-3}\\ \\=\frac{2+7}{2-3}\\ \\=\frac{9}{-1}\\ \\=-9\)

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.

\(\displaystyle \\ \lim_{x \to 3}\frac{x^2-6x+9}{x^4-81}\\ \\=\lim_{x \to 3}\frac{(x-3)(x-3)}{(x^2+9)(x^2-9)}\\ \\=\lim_{x \to 3}\frac{(x-3)(x-3)}{(x^2+9)(x+3)(x-3)}\\ \\=\lim_{x \to 3}\frac{x-3}{(x^2+9)(x+3)}\\ \\=\frac{3-3}{(3^2+9)(3+3)}\\ \\=\frac{0}{(9+9)(6)}\\ \\=\frac{0}{18*6}\\ \\=\frac{0}{108}=0\)

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=9; 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.

\(\displaystyle \\\lim_{x \to 9}\frac{x^2-9x}{x^2-8x-9}\\ \\=\lim_{x \to 9}\frac{x(x-9)}{(x-9)(x+1)}\\ \\=\lim_{x \to 9}\frac{x}{x+1}\\ \\=\frac{9}{9+1}=\frac{9}{10}\)

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.

\(\displaystyle \\\lim_{x \to -3}\frac{x^2-9}{x^2+7x+12}\\ \\=\lim_{x \to -3}\frac{(x+3)(x-3)}{(x+3)(x+4)}\\ \\=\lim_{x \to -3}\frac{x-3}{x+4}\\ \\=\frac{-3-3}{-3+4}\\ \\=\frac{-6}{1}\\ \\=-6\)

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=0; 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.

\(\displaystyle \\ \lim_{x \to 0}\frac{x^3+x^2}{x^4+x^3}\\ \\=\lim_{x \to 0}\frac{x^2(x+1)}{x^3(x+1)}\\ \\=\lim_{x \to 0}\frac{1}{x}\\ \\=\frac{1}{0}\Rightarrow DNE\)

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 not equal zero when x=2; so we proceed to insert the value of x into the entire equation.

\(\displaystyle \lim_{x \to 2}\frac{x-2}{x+2}=\frac{2-2}{2+2}=\frac{0}{4}=0\)