<s>[INST] Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
\right]
    = \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
    \left( 1 - \frac{1}{n} \right)^{2n}
    \left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
    -
    \left( 1 - \frac{2}{n} \right)^{n}
    \left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
  - \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
    \sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
  - \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
 [/INST] I would do the transform 
$$\left(1-\frac{1}{n}\right)^{2n}=e^{2n\log\left(1-\frac{1}{n}\right)}$$
then use the second order Taylor expansion
$$\log(1+x)\approx x-\frac{x^2}{2}$$
and similarly for the other term, obtaining
$$
n\left(e^{-2-\frac{1}{n}}-e^{-2-\frac{2}{n}}\right)=e^{-2-\frac{2}{n}}\cdot\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}
$$
</s>