Limit $\lim_{x\to0} \left(\frac1e(1+x)^{1/x}\right) ^{1/\sin x}$ The question goes like this 
Limit 
$$\lim_{x\to0} \left(\frac{(1+x)^{1/x}}{e}\right)^{1/\sin x}$$
Any help is welcome .
 A: Shortcuts are possible, but one very standard approach to limits of this kind is take the logarithm. Let $L$ be the desired limit. Then
$$\begin{align*}
\ln L&=\ln\lim_{x\to 0}\left(\frac{(1+x)^{1/x}}e\right)^{1/\sin x}\\\\
&=\lim_{x\to 0}\ln\left(\left(\frac{(1+x)^{1/x}}e\right)^{1/\sin x}\right)\\\\
&=\lim_{x\to 0}\left(\frac1{\sin x}\cdot\ln\frac{(1+x)^{1/x}}e\right)\\\\
&=\lim_{x\to 0}\frac{\ln(1+x)^{1/x}-1}{\sin x}\\\\
&=\lim_{x\to 0}\frac{\ln(1+x)-x}{x\sin x}
\end{align*}$$
where the interchange of the log and the limit is possible because the log is a continuous function. The limit succumbs easily to a couple of applications of l’Hospital’s rule.
Don’t forget, though, that after you’ve found $\ln L$, you’ll have to exponentiate to get $L$ itself.
A: If we assume the limit exists and is equal to $L$, then
$$\log{L} = \frac{(1/x)\log{(1+x)}-1}{\sin{x}} $$
The easiest way forward is to Taylor expand numerator and denominator - numerator should be expanded to the $x^2$ term.  The end result is
$$L = \frac{1}{\sqrt{e}}$$
A: Let $$f(x)=\left(\frac{(1+x)^{1/x}}{e}\right)^{1/\sin x}$$ Then $$\lim_{x\rightarrow 0}\ln f(x)=\lim_{x\rightarrow 0}\frac{1}{\sin x}\left(\frac{\ln (1+x)}{x}-1\right)=\lim_{x\rightarrow 0}\frac{1}{\sin x}\left(-\frac{x}{2}+\frac{x^2}{3}-\cdots\right)=-\frac{1}{2}$$
Since $\ln(\cdot)$ is continuous, we get $$\lim_{x\rightarrow 0}f(x)=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$$
A: First note that by definition of exponentiation
$$ \left(\frac{(1+x)^{1/x}}{e}\right)^{1/\sin x}
=\exp\left(\frac1{\sin x}\ln\frac{(1+x)^{1/x}}{e}\right)
=\exp\left(\frac1{\sin x}\left(\frac1x\ln(1+x)-1\right)\right)
$$
Now you can work your way from inside out using l'Hopital (or Taylor if you prefer)
A: Hints:
$$\left(\frac{(1+x)^{1/x}}e\right)^{1/\sin x}=e^{\frac1{\sin x}\left(\log(1+x)^{1/x}-1\right)}$$
So it suffices to evaluate the limit of the exponent above (why?). Note that
$$(1+x)^{1/x}=e^{\frac1x\log(1+x)}\xrightarrow[x\to 0]{}e\;,\;\;\text{by l'Hospital's rule, so :}$$
$$\lim_{x\to 0}\frac{\log(1+x)^{1/x}-1}{\sin x}\stackrel{\text{l'Hospital}}=\lim_{x\to 0}\left(-\frac1{x^2}\log(1+x)+\frac1{x(1+x)}\right)\cdot\frac1{\cos x}=$$
$$=\lim_{x\to 0}\frac{x-(1+x)\log(1+x)}{x^2(1+x)\cos x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{1-\log(1+x)-1}{2x(1+x)\cos x+x^2\cos x-x^2(1+x)\sin x}\stackrel{\text{l'H}}=$$
$$=\lim_{x\to 0}-\frac1{(1+x)\left(2\cos x+6x\cos x-4x(1+x)\sin x-2x^2\sin x-x^2(1+x)\cos x\right)}=$$
$$=-\frac1{1\cdot2}=-\frac12$$
and thus, finally,  the limit is ...
