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.

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:
$$\lim_{n\to\infty}\left(1-{a_n\over n}\right)^n=\exp\bigl(-e^2\bigr)\ .$$
