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Question Number 142668 by mathdanisur last updated on 03/Jun/21

lim_(n→∞) (((n+6)/n))^(6/n) = ?

$$\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}+\mathrm{6}}{{n}}\right)^{\frac{\mathrm{6}}{{n}}} =\:? \\ $$

Commented by iloveisrael last updated on 04/Jun/21

 lim_(x→∞)  (((n+6)/n))^(6/n)  =   e^(lim_(x→∞) (((n+6)/n)−1).(6/n))  =   e^(lim_(x→∞) (((36)/n^2 )))  = e^0  = 1

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}+\mathrm{6}}{{n}}\right)^{\frac{\mathrm{6}}{{n}}} \:= \\ $$$$\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}+\mathrm{6}}{{n}}−\mathrm{1}\right).\frac{\mathrm{6}}{{n}}} \:= \\ $$$$\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{36}}{{n}^{\mathrm{2}} }\right)} \:=\:{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by ajfour last updated on 03/Jun/21

ln L=lim_(n→∞) (6/n)ln (1+(6/n))=0  L=1.

$$\mathrm{ln}\:{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6}}{{n}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{6}}{{n}}\right)=\mathrm{0} \\ $$$${L}=\mathrm{1}. \\ $$

Answered by Ankushkumarparcha last updated on 03/Jun/21

Answer: e

$${Answer}:\:{e} \\ $$

Answered by mr W last updated on 03/Jun/21

lim_(n→∞) (((n+6)/n))^(6/n)   =lim_(n→∞) [(1+(1/(n/6)))^(n/6) ]^((36)/n^2 )   =e^0 =1

$$\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}+\mathrm{6}}{{n}}\right)^{\frac{\mathrm{6}}{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{{n}}{\mathrm{6}}}\right)^{\frac{{n}}{\mathrm{6}}} \right]^{\frac{\mathrm{36}}{{n}^{\mathrm{2}} }} \\ $$$$={e}^{\mathrm{0}} =\mathrm{1} \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

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