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

\(\displaystyle \lim_{n \to -5}\frac{n^2+4n-5}{n^2+7n+10}=\lim_{n \to -5}\frac{(n+5)(n-1)}{(n+5)(n+2)} \\ \\ =\lim_{n \to -5}\frac{n-1}{n+2}\\\\=\frac{-5-1}{-5+2}=\frac{-6}{-3}=2\)

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

\(\displaystyle \lim_{n \to -5}(5+n)(6-n)=(5+(-5))(6-(-5))=(5-5)(6+5)=0\cdot11=0\)

The limiting situation in this equation would be the denominator. Plug the value that \(\displaystyle n\) is approaching into the denominator to see if the denominator will equal \(\displaystyle 0\). In this question, the denominator will equal zero when \(\displaystyle n=6\); so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of \(\displaystyle n\) into the remaining equation.

\(\displaystyle \lim_{n \to 6}\frac{3n^2+15n-18}{6n^2+12n-288}=\lim_{n \to 6}\frac{3(n^2-5n-6)}{6(n^2+2n-48)} \\ \\ =\lim_{n \to 6}\frac{3(n-6)(n+1)}{6(n-6)(n+8)}=\lim_{n \to 6}\frac{n+1}{2(n+8)} \\ \\ =\frac{6+1}{2(6+8)}=\frac{7}{2(14)}=\frac{1}{2\cdot2}=\frac{1}{4}\)

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

\(\displaystyle \lim_{n \to 6}n^2+n=(6^2)+6=36+6=42\)

The limiting situation in this equation would be the denominator. Plug the value that \(\displaystyle n\) is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when \(\displaystyle n=-6\); so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of \(\displaystyle n\) into the remaining equation.

\(\displaystyle \lim_{n \to -6}\frac{n^2+9n+18}{n^2-36}=\lim_{n \to -6}\frac{(n+6)(n+3)}{(n+6)(n-6)} \\ \\ =\lim_{n \to -6}\frac{n+3}{n-6}=\frac{-6+3}{-6-6} \\ \\ =\frac{-3}{-12}=\frac{1}{4}\)

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

\(\displaystyle \lim_{n \to -6}3n^2+2n-6=3(-6)^2+2(-6)-6=3\cdot36-12-6=108-18=90\)

In order to evaluate the limit, we must see whether the limit is right or left sided. The negative sign exponent on the \(\displaystyle 3\) indicates we are approaching \(\displaystyle 3\) from the left side, or with values slightly less than \(\displaystyle 3\). The part of the piecewise function corresponding to these values is the first function, and evaluated at \(\displaystyle 3\), we get \(\displaystyle e^3\)

To evaluate the limit as \(\displaystyle x\) approaches infinity, we must first take out a term consisting of the highest power term divided by itself (\(\displaystyle 1\), essentially:

\(\displaystyle \lim_{x\rightarrow \infty }\frac{x^4}{x^4}(\frac{8+4x^{-1}}{3+x^{-1}+2x^{-2}})\)

When the term we created becomes \(\displaystyle 1\), and the terms with negative exponents become zero (as \(\displaystyle x\) approaches infinity, the fraction with infinity in the denominator equals zero), we are left with our answer, \(\displaystyle \frac{8}{3}\)

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

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

The limiting situation in this equation would be the denominator. Plug the value that \(\displaystyle n\) is approaching into the denominator to see if the denominator will equal \(\displaystyle 0\). In this question, the denominator will equal zero when \(\displaystyle n=1\); so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of \(\displaystyle n\) into the remaining equation.

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