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Question-212075




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)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}+\sqrt{{n}}}−\sqrt{{n}}\right)\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{t}}}−\mathrm{1}}{\:\sqrt{{t}}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\sqrt{{t}}}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 28/Sep/24
thanks
$${thanks} \\ $$
Answered by mr W last updated on 28/Sep/24
=lim_(n→∞) ((√n)/( (√(n+(√n)))+(√n)))  =lim_(n→∞) (1/( (√(1+(1/( (√n)))))+1))  =(1/(1+1))=(1/2)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{n}}}{\:\sqrt{{n}+\sqrt{{n}}}+\sqrt{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{n}}}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 28/Sep/24
thanks mr W
$${thanks}\:{mr}\:{W} \\ $$
Answered by mehdee7396 last updated on 28/Sep/24
lim_(n→∞)  ((n+(√n)−n)/( (√(n+(√n)))+(√n)))=  =lim_(n→∞)  ((√n)/( 2(√n)))=(1/2) ✓
$${lim}_{{n}\rightarrow\infty} \:\frac{{n}+\sqrt{{n}}−{n}}{\:\sqrt{{n}+\sqrt{{n}}}+\sqrt{{n}}}= \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\sqrt{{n}}}{\:\mathrm{2}\sqrt{{n}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$

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