# How would you find this limit ? $$\displaystyle\lim_{{{n}\to\infty}}{\left({1}+{\ln{{\left({1}-{\frac{{{1}}}{{{2}{n}}}}\right)}}}\right)}^{{n}}$$

How would you find this limit ?
$\underset{n\to \mathrm{\infty }}{lim}{\left(1+\mathrm{ln}\left(1-\frac{1}{2n}\right)\right)}^{n}$
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sa3b4or9i9
We have
$\underset{n\to \mathrm{\infty }}{lim}{\left(1+\mathrm{ln}\left(1-\frac{1}{2n}\right)\right)}^{n}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\left(1+\mathrm{ln}\left(1-\frac{1}{2n}\left({\left({}^{\left\{\frac{1}{\mathrm{ln}\left(1-\frac{1}{2n}\right)}\right\}}\right)}^{n\mathrm{ln}\left(1-\frac{1}{2n}\right)}={e}^{\underset{n\to \mathrm{\infty }}{lim}n\mathrm{ln}\left(1-\frac{1}{2n}\right)}$
provided the limit in the exponent exists. (Since
$\underset{n\to \mathrm{\infty }}{lim}\left({\left(1+\mathrm{ln}\left(1-\frac{1}{2n}\right)\right)}^{\frac{1}{\mathrm{ln}\left(1-\frac{1}{2n}\right)}}\right)=\underset{x\to 0}{lim}{\left(1+x\right)}^{\frac{1}{x}}=e$ )
So it only remains to compute the limit from the exponent. We get
$\underset{n\to \mathrm{\infty }}{lim}n\mathrm{ln}\left(1-\frac{1}{2n}\right)=-{\frac{12}{lim}}_{n\to \mathrm{\infty }}\frac{\mathrm{ln}\left(1-\frac{1}{2n}\right)}{-\frac{1}{2n}}=-\frac{12}{}$
using $\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}=1$
So the original limit is ${e}^{-\frac{1}{2}}$.