Question Number 197407 by mathlove last updated on 16/Sep/23
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$
Answered by Rasheed.Sindhi last updated on 16/Sep/23
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}=\frac{\sqrt{\mathrm{1}+\mathrm{0}}}{\mathrm{1}+\mathrm{0}}=\mathrm{1} \\ $$
Commented by mathlove last updated on 16/Sep/23
$${thanks} \\ $$
Answered by MM42 last updated on 16/Sep/23
$${lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}}{{x}+\mathrm{1}}\:=\:{lim}_{{x}\rightarrow\infty} \:\frac{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}\: \\ $$$$={lim}_{{x}\rightarrow\infty} \:\frac{\mid{x}\mid}{{x}}\:=\begin{cases}{\mathrm{1}\:\:\:\:\:\:\:;\:\:{x}\rightarrow+\infty}\\{−\mathrm{1}\:\:\:;\:\:{x}\rightarrow−\infty}\end{cases} \\ $$$$\Rightarrow{lim}\:\:\::\:\:{not}\:\:\:{exist} \\ $$
Commented by MathematicalUser2357 last updated on 17/Sep/23
$$\mathrm{So}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mid{x}\mid}{{x}}\neq\mathrm{1} \\ $$
Commented by MM42 last updated on 18/Sep/23
$${lim}_{{x}\rightarrow+\infty} \:\frac{{x}}{{x}}=\mathrm{1}\:\:\&\:\:{lim}_{{x}\rightarrow−\infty} \:\frac{−{x}}{{x}}\:=−\mathrm{1} \\ $$