Exponential limit convergence for each $x$ I have
$f_n(x)=\left( 1+\frac{-e^{-x}}{n} \right)^n$, what about the convergence to $f(x)=e^{-e^{-x}}$?
is it true $\forall x?$ I say yes, but how can I show this? Is continuity of $f_n$ enough?
 A: If one knows the following Taylor series expansion, as $u \to 0$,
$$
\log(1+u)=u+O(u^2)
$$ then one may write, for any fixed real number $t$, as $n \to \infty$,
$$
\left( 1+\frac{t}{n} \right)^{n}=e^{\large n\log\left(1+\frac{t}{n}\right)}=e^{\large n\left(\frac{t}{n}+O\left(\frac{1}{n^2}\right)\right)}=e^{t+O\left(\frac{1}{n}\right)}
$$ which gives

$$
\lim_{n \to \infty}\left( 1+\frac{t}{n} \right)^{n}=e^t.
$$ 

The announced result is obtained with $t=-e^{-x}$.
A: It also works using L'Hopital's rule observing $$\lim_{n\rightarrow\infty}\left(1-\frac{e^{-x}}{n}\right)^{n}=e^{\lim_{n\rightarrow\infty}n\log\left(1-\frac{e^{-x}}{n}\right)}
 $$ and so $$\lim_{n\rightarrow\infty}n\log\left(1-\frac{e^{-x}}{n}\right)=-e^{-x}\lim_{n\rightarrow\infty}\frac{\log\left(1-\frac{e^{-x}}{n}\right)}{-\frac{e^{-x}}{n}}
 $$ $$=-e^{-x}\lim_{n\rightarrow\infty}\frac{e^{-x}}{n^{2}-e^{-x}n}\frac{n^{2}}{e^{-x}}=-e^{-x}.$$
A: Convergence is guaranteed for any $t=-e^{-x}$, as
$$\lim_{n\to\infty}\left(1+\frac tn\right)^n=\lim_{m\to\infty}\left(1+\frac 1m\right)^{mt}=\left(\lim_{m\to\infty}\left(1+\frac 1m\right)^m\right)^t=e^t.$$
No property of $f_n$ is required.
