<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] In the same spirit as Lucian, you can consider the general expansion for large values of $n$ $$ \left( 1 - \frac{a}{n} \right)^{b\,n}= \left(1-\frac{a^2 b}{2 n}+\frac{a^3 b (3 a b-8)}{24 n^2}-\frac{a^4 b
   \left(a^2 b^2-8 a b+12\right)}{48 n^3}+\cdots\right)e^{-a b} $$ Applied to your case $$\left( 1 - \frac{1}{n} \right)^{2n}=\left(1-\frac 1n-\frac1 {6n^2}+\cdots\right)e^{-2}$$ $$\left( 1 - \frac{2}{n} \right)^{n}=\left(1-\frac 2n-\frac2 {3n^2}+\cdots\right)e^{-2}$$ which finally make $$n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
\right]=\left( 1 +\frac{1}{2n}+\cdots \right)e^{-2}$$
In a more general manner,$$\left[
    \left( 1 - \frac{a}{n} \right)^{bn}
    - \left( 1 - \frac{b}{n} \right)^{an}
\right]=(a-b)\left(-\frac{a b }{2 n}+\frac{a b  (a+b) (3 a b-8)}{24 n^2}\right)e^{-ab}$$
</s>