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  • Find <img alt="\mathrm{\lim _{x \rightarrow \infty}\left[\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)\right]^x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%

Find \mathrm{\lim _{x \rightarrow \infty}\left[\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)\right]^x}
 

Option: 1

\mathrm{\frac{1}{\sqrt e}}


Option: 2

1


Option: 3

\mathrm{e}


Option: 4

0


Answers (1)

best_answer

By the Taylor expansion of \mathrm{\log (1+z)} in a neighbourhood of zero, when \mathrm{x \rightarrow+\infty} we have:
\mathrm{ x \log \left(1+\frac{1}{x}\right)-1=-\frac{1}{2 x}+\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right) }              (1)
so, exponentiating the previous identity,

\mathrm{ \frac{1}{e}\left(1+\frac{1}{x}\right)^x=1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right) }                                    (2)

and by raising both terms to the \mathrm{ x }-power:

\mathrm{ \lim _{x \rightarrow+\infty}\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)^x=\lim _{x \rightarrow+\infty}\left(1-\frac{1}{2 x}\right)^x=\frac{1}{\sqrt{e}} }    (3)

Hence option 1 is correct.


 

Posted by

jitender.kumar

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