Question Number 212075 by Spillover last updated on 28/Sep/24
Answered by Ghisom last updated on 28/Sep/24
![lim_(n→∞) ((√(n+(√n)))−(√n)) =lim_(t→0) (((√(1+(√t)))−1)/( (√t))) = [l′Ho^ pital] =lim_(t→0) (1/(2(√(1+(√t))))) =(1/2)](https://www.tinkutara.com/question/Q212079.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}+\sqrt{{n}}}−\sqrt{{n}}\right)\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{t}}}−\mathrm{1}}{\:\sqrt{{t}}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\sqrt{{t}}}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 28/Sep/24
$${thanks} \\ $$
Answered by mr W last updated on 28/Sep/24
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{n}}}{\:\sqrt{{n}+\sqrt{{n}}}+\sqrt{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{n}}}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 28/Sep/24
$${thanks}\:{mr}\:{W} \\ $$
Answered by mehdee7396 last updated on 28/Sep/24
$${lim}_{{n}\rightarrow\infty} \:\frac{{n}+\sqrt{{n}}−{n}}{\:\sqrt{{n}+\sqrt{{n}}}+\sqrt{{n}}}= \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\sqrt{{n}}}{\:\mathrm{2}\sqrt{{n}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$