Find:

Solution (a) We find this limit by dividing each term of the numerator and the denominator by the term with the highest power of $x$ that appears in the denominator, in this case, $x^{2}$. Then $$\begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{3x^{2}\,{-}\,2x\,{+}\,8}{x^{2}+1}\hspace{-2.8pc}\underset{\underset{\color{#0066A7}{\raise-12pt\scriptsize\hbox{Divide the numerator and }}}{\color{#0066A7}{\uparrow}}}{=}\hspace{-3pc}\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{ 3x^{2}\,{-}\,2x\,{+}\,8}{x^{2}}}{\dfrac{x^{2}+1}{x^{2}}}&=&\lim\limits_{x\rightarrow \infty }\dfrac{3\,{-}\,\dfrac{2}{x}\,{+}\,\dfrac{8}{x^{2}}}{1+\dfrac{1}{x^{2}}}\hbox{ }\hspace{-2.5pc} \underset{\raise-5pt\underset{\color{#0066A7}{\raise-13pt\scriptsize\hbox{Limit of a Quotient}}}{\color{#0066A7}{\uparrow}}}{=}\hspace{-2pc}\dfrac{\lim\limits_{x\rightarrow \infty }\left[ 3\,{-}\,\dfrac{2}{x}\,{+}\,\dfrac{8}{x^{2}}\right] }{\lim\limits_{x\rightarrow \infty }\left[ 1+\dfrac{1 }{x^{2}}\right] }\\[-1pc] \hspace{-8pc}{\color{#0066A7}{\scriptsize\hbox{denominator by \(x^{2}\)}}} \\ &&\\[-21pc] &=&\dfrac{\lim\limits_{x\rightarrow \infty }3-\lim\limits_{x\rightarrow \infty }\dfrac{2}{x}+\lim\limits_{x\rightarrow \infty }\dfrac{8}{x^{2}}}{\lim\limits_{x\rightarrow \infty }1+\lim\limits_{x\rightarrow \infty }\dfrac{1}{x^{2}}}=\dfrac{ 3-2\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}+8\left(\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}\right)^2}{1+\left(\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}\right)^2}\\[3pt] &=&\dfrac{3-0+0}{1+0}=3 \end{eqnarray*}$$

(b) $$\lim\limits_{x\rightarrow -\infty }\dfrac{4x^{2}-5x}{ x^{3}+1}\hspace{-3pc}\raise-4pt\underset{\underset{\color{#0066A7}{\scriptsize\hbox{Divide the numerator and}}}{\color{#0066A7}{\uparrow}}} {=}\hspace{-2.9pc}\lim\limits_{x\rightarrow -\infty }\dfrac{\dfrac{4x^{2}-5x}{x^{3}}}{\dfrac{x^{3}+1}{x^{3}}} =\lim\limits_{x\rightarrow -\infty }\dfrac{\dfrac{4}{x}-\dfrac{5}{x^{2}}}{1+ \dfrac{1}{x^{3}}}=\dfrac{\lim\limits_{x\rightarrow -\infty }\left( \dfrac{4}{ x}-\dfrac{5}{x^{2}}\right) }{\lim\limits_{x\rightarrow -\infty }\left( 1+ \dfrac{3}{x^{3}}\right) }=\dfrac{0}{1}=0\\[-20pt] \hspace{-15pc}{\raise0pt\color{#0066A7}{\scriptsize\hbox{denominator by \(x^{3}\)}}}$$