Asymptotic behavior of the expression: $(1-\frac{\ln n}{n})^n$ when $n\rightarrow\infty$ The well known results states that:
$\lim_{n\rightarrow \infty}(1-\frac{c}{n})^n=(1/e)^c$ for any constant $c$.
I need the following limit: $\lim_{n\rightarrow \infty}(1-\frac{\ln n}{n})^n$.
Can I prove it in the following way? Let $x=\frac{n}{\ln n}$, then we get: $\lim_{n\rightarrow \infty}(1-\frac{\ln}{n})^n=\lim_{x\rightarrow \infty}(1-\frac{1}{x})^{x\ln n}=(1/e)^{\ln n}=\frac{1}{n}$.
So,  $\lim_{n\rightarrow \infty}(1-\frac{\ln}{n})^n=\frac{1}{n}$.
I see that this is wrong to have an expression with $n$ after the limit. But how to show that the asymptotic behavior is $1/n$?
Thanks!

The idea you used could have worked. Exactly as you wrote, we have 
$$\left(1-\frac{\ln n}{n}\right)^n=\left(\left(1-\frac{1}{x}\right)^x\right)^{\ln n},$$
where $x=\frac{n}{\ln n}$. 
Note that $x\to \infty$ as $n\to\infty$. Note also that $\ln n=g(x)$ for some function $g(x)$ such that $g(x)\to\infty$ as $x\to \infty$. Our limit is 
$$\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{g(x)}.$$
Since $\left(1-\frac{1}{x}\right)^x$ approaches $\frac{1}{e}$, and $g(x)\to\infty$, our limit is $0$.
Remark:  Your impossible answer $\frac{1}{n}$ contained a large kernel of truth. When $n$ is large, $(1-1/x)^x$ is close to $1/e$, so the original expression is close to $1/n$.  And of course the quantity $1/n$ approaches $0$.  
Let $E(n)$ be the original expression. We can adapt your argument to show that $\lim_{n\to\infty} nE(n)=1$, which proves in a very informative way that $E(n)$ has limit $0$, by giving quite exact information about the rate of approach to $0$.  
