Find $\lim_{x\to \infty}{[({1\over e}(1+{1\over x})^x)]^x}$. Find $\lim_{x\to \infty}{[({1\over e}(1+{1\over x})^x)]^x}$. 
I have been trying for hours using the continuity of $e$ and using L'Hopital rule but it gets really scattered and ugly. I am in despaire. I would really, truly, appreciate your help. 
 A: By the Taylor expansion of $\log(1+z)$ in a neighbourhood of zero, when $x\to +\infty$ we have:
$$ x\log\left(1+\frac{1}{x}\right)-1 = -\frac{1}{2x}+\frac{1}{3x^2}+O\left(\frac{1}{x^3}\right)\tag{1}$$
so, exponentiating the previous identity,
$$\frac{1}{e}\left(1+\frac{1}{x}\right)^x = 1-\frac{1}{2x}+O\left(\frac{1}{x^2}\right) \tag{2}$$
and by raising both terms to the $x$-power:
$$ \lim_{x\to +\infty}\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)^x = \lim_{x\to +\infty}\left(1-\frac{1}{2x}\right)^x = \color{red}{\frac{1}{\sqrt{e}}}.\tag{3}$$
A: Note $f(x) = {[({1\over e}(1+{1\over x})^x)]^x}$, then $\log(f(x)) = - x \log (e) + x \log \left( \left( 1+\frac{1}{x}\right)^x \right) = -x + x^2 \log\left( 1+\frac{1}{x}\right) = x^2 \left( \log\left( 1+\frac{1}{x}\right) - \frac{1}{x}\right) $ and as when $x$ then to $+\infty$ the quantity $\log\left( 1+\frac{1}{x}\right) - \frac{1}{x}$ is equivalent to $-\frac{1}{2}\left(\frac{1}{x}\right)^2$ (develop $\log(1+u)$ at the second order in $u$ as $u\to 0$) you find that $\log(f(x)) \to -\frac{1}{2}$, and taking the $\exp$ gives you that $f(x) \to e^{-\frac{1}{2}}$.
A: Hint 1: try taylor, which is usually more doable than L'Hospital.
Hint 2: Try the logarithm. Calculate $\displaystyle\lim_{x\to\infty}\ln [\frac{1}{e}(1+\frac{1}{x})^x]^x$ first
$$\displaystyle\lim_{x\to\infty}\ln [\frac{1}{e}(1+\frac{1}{x})^x]^x=\lim_{x\to\infty}[x^2\ln(1+\frac{1}{x})-x]$$
Since $\frac{1}{x}\to 0$
$$\ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2}\frac{1}{x^2}+o(\frac{1}{x^2})$$
Therefore
$$x^2\ln(1+\frac{1}{x})-x=-\frac{1}{2}+o(\frac{1}{x})$$
Or
$$\lim_{x\to\infty}[x^2\ln(1+\frac{1}{x})-x]=\lim_{x\to\infty}[-\frac{1}{2}+o(\frac{1}{x})]=\lim_{x\to\infty}\ln [\frac{1}{e}(1+\frac{1}{x})^x]^x=-\frac{1}{2}$$
Which yields
$$\lim_{x\to\infty}[\frac{1}{e}(1+\frac{1}{x})^x]^x=e^{-\frac{1}{2}}$$
