How I can find this limit? 
If $a_n=(1+\frac{2}{n})^n$ , then find $$\lim_{n \to \infty}(1-\frac{a_n}{n})^n$$.

Trial: Can I use $$\lim_{n \to \infty}a_n=e^2$$ Again $$\lim_{n \to \infty}(1-\frac{a_n}{n})^n=\exp(-e^2)$$ Please help.


If $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ uniformly on $E$, $f$ is continuous at $a\in E$ and $a_n\xrightarrow[n\to\infty]{}a$ with $a_n\in E \ \ \forall  \ n \in \mathbb N$, then $$f_n(a_n)\xrightarrow[n\to\infty]{}f(a).$$

Indeed, let $\epsilon>0$ and $\delta>0$ such that $|f(x)-f(a)|<\dfrac{\epsilon}{2}$ whenever $|x-a|<\delta$. Let $n_0\in\mathbb N$ such that $|f_n(x)-f(x)|<\dfrac{\epsilon}{2} \text{ and }  |a_n-a|<\delta, \ \forall x\in E, \ n\geq n_0$. Then for $n\geq n_0$:
$$ |f_n(a_n)-f(a)|\leq |f_n(a_n)-f(a_n)|+|f(a_n)-f(a)|<\epsilon \ . $$

Now apply the previous to $f_n(x)=(1-\frac xn)^n$ (see this) and $a_n=(1+\frac2n)^n$.
