Question Number 170812 by MathsFan last updated on 31/May/22
$$\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{x}}+\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{horizontal}}\:\boldsymbol{\mathrm{asymptote}} \\ $$
Answered by Mathspace last updated on 31/May/22
$${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow−\infty} \frac{{x}+\mathrm{1}}{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }}} \\ $$$$={lim}_{{x}\rightarrow−\infty} \frac{{x}+\mathrm{1}}{−{x}\sqrt{\mathrm{1}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }}} \\ $$$$=−\mathrm{1}\:\Rightarrow{y}=−\mathrm{1}\:{is}\:{assymtote} \\ $$$${at}\:−\infty \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow+\infty} \frac{{x}+\mathrm{1}}{{x}\sqrt{\mathrm{1}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }}} \\ $$
Commented by Mathspace last updated on 31/May/22
$$=\mathrm{1}\:\Rightarrow{y}=\mathrm{1}\:{is}\:{assymptote}\:{at}\:+\infty \\ $$