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.

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!
