Dealing with indeterminate forms of the $1^\infty $ kind $$\lim\limits_{x→{\frac π{2}}^-}\left(\frac {2x}{\pi}\right)^ {\tan x}$$  and
$$\lim\limits_{n→\infty} \left(1+ \frac {1}{n}\right)^n$$  
could anyone provide some hints? how to start. (with possibility of using L'Hop rule. )
 A: We have $$ \lim_{x\rightarrow\frac{\pi}{2}^{-}}\left(\frac{2x}{\pi}\right)^{\tan\left(x\right)}=\exp\left(\lim_{x\rightarrow\frac{\pi}{2}^{-}}\tan\left(x\right)\left(\log\left(\frac{2}{\pi}\right)+\log\left(x\right)\right)\right)=
 $$ $$=\exp\left(\lim_{x\rightarrow\frac{\pi}{2}^{-}}\sin\left(x\right)\lim_{x\rightarrow\frac{\pi}{2}^{-}}\frac{\left(\log\left(\frac{2}{\pi}\right)+\log\left(x\right)\right)}{\cos\left(x\right)}\right)=\exp\left(\lim_{x\rightarrow\frac{\pi}{2}^{-}}\frac{\left(\log\left(\frac{2}{\pi}\right)+\log\left(x\right)\right)}{\cos\left(x\right)}\right)
 $$ and now we can apply De L'Hopital's rule to get $$=\exp\left(\lim_{x\rightarrow\frac{\pi}{2}^{-}}-\frac{1}{x\sin\left(x\right)}\right)=e^{-2/\pi}.
 $$ The second is the definition of $e$. If you prefer, $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}=\exp\left(\lim_{n\rightarrow\infty}n\log\left(1+\frac{1}{n}\right)\right)=\exp\left(\lim_{n\rightarrow\infty}\frac{\log\left(1+\frac{1}{n}\right)}{\frac{1}{n}}\right)=e
 $$ using $$\lim_{x\rightarrow0}\frac{\log\left(1+x\right)}{x}=1.
 $$
A: Taking logarithm and applying L'Hopital's rule, we have
 $$\lim\limits_{x\to{\frac π{2}}^-}\ln\left[\left(\frac {2x}{\pi}\right)^ {\tan x}\right]=\lim\limits_{x\to{\frac π{2}}^-}\tan{x}\cdot\ln{\left(\frac {2x}{\pi}\right)}=\lim\limits_{x\to{\frac π{2}}^-}\frac{\ln{\left(\frac {2x}{\pi}\right)}}{\cot{x}}=\left[\frac{0}{0}\right] =\\
=\lim\limits_{x\to{\frac π{2}}^-}\frac{\frac{1}{x}}{-\frac{1}{\sin^2{x}}}=-\lim\limits_{x\to{\frac π{2}}^-}\frac{\sin^2{x}}{x}=-\frac{2}{\pi},$$
therefore,
$$\lim\limits_{x→{\frac π{2}}^-}\left(\frac {2x}{\pi}\right)^ {\tan x}=e^{-\frac{2}{\pi}}.$$
A: for the second limit:
from mac-lauren series we have: $x\to 0\Rightarrow \ln(1+x)\approx x+\frac{x^2}{2}$ so:
$$\lim\limits_{n→\infty} \left(1+ \frac {1}{n}\right)^n=\lim_{n\to +\infty}e^{n\ln(1+\frac{1}{n})}=\lim_{n\to +\infty}e^{n(\frac{1}{n}+\frac{1}{2n^2})}=\lim_{n\to\infty}e^{1+\frac{1}{2n}}=e^1=e$$
