Finding Limits at Infinity

Find:

1. \(\lim\limits_{x\rightarrow \infty }\) \(\dfrac{3x^{2}-2x+8}{x^{2}+1}\)
2. \(\lim\limits_{x\rightarrow -\infty }\dfrac{4x^{2}-5x}{x^{3}+1}\)

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 \(x^{3}\)}}} \]