Compute the limit $\lim_{n\to\infty}{(\sqrt[n]{e}-\frac{2}{n})^n}$ How can I compute this limit of the sequence?
 $$\lim_{n\to\infty}{(\sqrt[n]{e}-\frac{2}{n})^n}$$
 A: This depends on the tools you master, but if you are aware that $$\sqrt[n]{\mathrm e}=\mathrm e^{1/n}=1+1/n+o(1/n),$$ then you can write that $$(\sqrt[n]{\mathrm e}-2/n)^n=(1-1/n+o(1/n))^n\to \mathrm e^{-1}.$$
A: With the substitution $x=1/t$ we have
$$
\lim_{x\to\infty}(e^{1/x}-2/x)^x=
\lim_{t\to0^+}(e^t-2t)^{1/t}
$$
The limit of the logarithm of the expression $(e^t-2t)^{1/t}$ is
$$
\lim_{t\to0^+}\frac{\log(e^t-2t)}{t}
\overset{\mathrm{(H)}}{=}
\lim_{t\to0^+}\frac{e^t-2}{e^t-2t}=-1
$$
Without l'Hôpital, recall that $\log(e^t-2t)=\log(1-t+o(t))=-t+o(t)$.
A: Another way you can go is:
$$\begin{align}
\lim_{n\to\infty}\log\left(\sqrt[n]e-\frac2n\right)^n & =\lim_{n\to\infty}{\log\left(e^{1/n}-\frac2n\right)\over\frac1n}\\
& = \lim_{n\to\infty}{\left(-\frac{1}{n^2}e^{1/n}+\frac{2}{n^2}\right)/\left(e^{1/n}-\frac{2}{n}\right)\over -\frac{1}{n^2}}
\end{align}$$,
by L'Hospital's Rule. Now multiply numerator and denominator by $-n^2$, to get 
$${e^{1/n}-2\over e^{1/n}-\frac2n}$$
and it should be fairly easy.
