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.
 A: Due to 
$$(1-\frac{a_n}{n})^n=\left[\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}\right]^{\frac{-a_n}{n}n}=\left[\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}\right]^{-a_n}.$$
$$\lim_{n\to\infty}\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}=e$$
and $$\lim_{n\to \infty}(-a_n)=-e^2$$
Let $A_n=\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}$ and $B_n=-a_n$, by the "claim" below, you can get the result!
A: 
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 $a_n=(1+\frac2n)^n$.
A: This is a very nice problem; but it does not file under "composite limits", i.e., limits of the form $\lim_{x\to \xi} f\bigl(g(x)\bigr)$.
It is agreed that ${\displaystyle\lim_{n\to\infty} a_n=\lim_{n\to\infty}\left(1+{2\over n}\right)^n = e^2}$.
The function $u\mapsto {\rm Log}(1-u)$ is analytic for $|u|<1$. Therefore "by Taylor's Theorem" there is a function $u\mapsto g(u)$, analytic for $|u|<1$, such that
$$\log(1-u)=-u + u^2 g(u)\ .\qquad(1)$$
It follows that there is an $M>0$ such that 
$$\bigl|g(u)\bigr|\leq M\qquad\left(|u|\leq{1\over2}\right)\ .$$
If $x$ and $y$ are arbitrary real numbers with $0<|xy|<1$ then we get from $(1)$ that
$${\log(1-x y)\over x}=-y + x y^2 g(xy)\ .$$
Now put $x:={1\over n}$, $\ y:=a_n$. Then the last equation says
$$n\ \log\left(1-{a_n\over n}\right)=-a_n +{1\over n} a_n^2\> g\left({a_n\over n}\right)\ ,\qquad(2)$$
whereby this holds as soon as $n$ is large enough to make $|a_n|<n$.
Now let $n\to\infty$. Then $a_n\to e^2$ and $\ {a_n\over n}\to 0$. Therefore the RHS of $(2)$ converges to $-e^2$. So does the LHS, and we get what we have conjectured all along:
$$\lim_{n\to\infty}\left(1-{a_n\over n}\right)^n=\exp\bigl(-e^2\bigr)\ .$$
