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.

\(\displaystyle \lim_{n \to 2}\frac{n^2-n-2}{n^2-5n+6}=\lim_{n \to 2}\frac{(n-2)(n+1)}{(n-2)(n-3)}=\lim_{n \to 2}\frac{n+1}{n-3}=\frac{2+1}{2-3}=\frac{3}{-1}=-3\)

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.

\(\displaystyle \lim_{n \to -3}\frac{n^2+2n-3}{n^2+4n+3}=\lim_{n \to -3}\frac{(n+3)(n-1)}{(n+3)(n+1)}=\lim_{n \to -3}\frac{n-1}{n+1}=\frac{-3-1}{-3+1}=\frac{-4}{-2}=2\)

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:

\(\displaystyle \lim_{n \to 1}n^2+3n+3=(1)^2+3(1)+3=1+3+3=7\)

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:

\(\displaystyle \lim_{n \to 2}n^2+\sqrt{n}=(2)^2+\sqrt{2}=4+\sqrt{2}\)

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:

\(\displaystyle \lim_{n \to -2}5n-4=5(-2)-4=-10-4=-14\)

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:

\(\displaystyle \lim_{n \to -3}15+5n=15+5(-3)=15-15=0\)

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=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 n into the remaining equation.

\(\displaystyle \lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}=\lim_{n \to 1}\frac{(n-1)(n-1)}{(n-1)(n+1)}=\lim_{n \to 1}\frac{n-1}{n+1}=\frac{1-1}{1+1}=\frac{0}{2}=0\)

This particular question is asking us to find a one sided limit of the functions depicted by the graph. Since there is a plus sign in the exponent on the zero, this tells us that we are looking for a right handed limit. In other words, we will loook at the function values for x values that are slightly larger, or to the right of 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.

This particular question is asking us to find a one sided limit of the functions depicted by the graph. Since there is a negative sign in the exponent of the zero, this tells us that we are looking for a left handed limit. In other words, we will loook at the function values for x values that are slightly smaller, or to the left of zero.

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

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:

\(\displaystyle \lim_{n \to -1}n^3+n^4-n^5=(-1)^3+(-1)^4-(-1)^5=-1+1+1=1\)