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!

No, it is incorrect. The answer is 0.
Since $\lim_{n\to\infty}(1-\frac{\ln n}{n})^{\frac{n}{\ln n}}=\frac{1}{e}<1$ and 
$\lim_{n\to\infty}\ln n=+\infty$, we have that 
$\lim_{n\to\infty}(1-\frac{\ln n}{n})^n=\lim_{n\to\infty}((1-\frac{\ln n}{n})^{\frac{n}{\ln n}})^{\ln n}=0$.
