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Question Number 197407 by mathlove last updated on 16/Sep/23

lim_(x→∞) ((√(x^2 +1))/(x+1))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$

Answered by Rasheed.Sindhi last updated on 16/Sep/23

lim_(x→∞) ((√(x^2 +1))/(x+1))=lim_(x→∞) ((x(√(1+(1/x^2 ))))/(x(1+(1/x)))) =lim_(x→∞) ((√(1+(1/x^2 )))/(1+(1/x)))=((√(1+0))/(1+0))=1

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}=\frac{\sqrt{\mathrm{1}+\mathrm{0}}}{\mathrm{1}+\mathrm{0}}=\mathrm{1} \\ $$

Commented by mathlove last updated on 16/Sep/23

thanks

$${thanks} \\ $$

Answered by MM42 last updated on 16/Sep/23

lim_(x→∞) ((√(x^2 (1+(1/x^2 ))))/(x+1)) = lim_(x→∞) ((∣x∣(√(1+(1/x^2 ))))/(x(1+(1/x)))) =lim_(x→∞) ((∣x∣)/x) = { ((1 ; x→+∞)),((−1 ; x→−∞)) :} ⇒lim : not exist

$${lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}}{{x}+\mathrm{1}}\:=\:{lim}_{{x}\rightarrow\infty} \:\frac{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}\: \\ $$$$={lim}_{{x}\rightarrow\infty} \:\frac{\mid{x}\mid}{{x}}\:=\begin{cases}{\mathrm{1}\:\:\:\:\:\:\:;\:\:{x}\rightarrow+\infty}\\{−\mathrm{1}\:\:\:;\:\:{x}\rightarrow−\infty}\end{cases} \\ $$$$\Rightarrow{lim}\:\:\::\:\:{not}\:\:\:{exist} \\ $$

Commented by MathematicalUser2357 last updated on 17/Sep/23

So lim_(x→∞) ((∣x∣)/x)≠1

$$\mathrm{So}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mid{x}\mid}{{x}}\neq\mathrm{1} \\ $$

Commented by MM42 last updated on 18/Sep/23

lim_(x→+∞) (x/x)=1 & lim_(x→−∞) ((−x)/x) =−1

$${lim}_{{x}\rightarrow+\infty} \:\frac{{x}}{{x}}=\mathrm{1}\:\:\&\:\:{lim}_{{x}\rightarrow−\infty} \:\frac{−{x}}{{x}}\:=−\mathrm{1} \\ $$

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