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!
 A: According to the comments, your real aim is to prove that $x_n=n\left(1-\frac{\log n}n\right)^n$ has a non degenerate limit. 
Note that $\log x_n=\log n+n\log\left(1-\frac{\log n}n\right)$ and that $\log(1-u)=-u+O(u^2)$ when $u\to0$ hence $n\log\left(1-\frac{\log n}n\right)=-\log n+O\left(\frac{(\log n)^2}n\right)$ and $\log x_n=O\left(\frac{(\log n)^2}n\right)$.
In particular, $\log x_n\to0$, hence $x_n\to1$, that is,
$$
\left(1-\frac{\log n}n\right)^n\sim\frac1n.
$$
Edit: In the case at hand, one knows that $\log(1-u)\leqslant-u$ for every $u$ in $[0,1)$. Hence $\log x_n\leqslant0$ and, for every $n\geqslant1$,
$$
\left(1-\frac{\log n}n\right)^n\leqslant\frac1n.
$$
A: 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 $0$.  
A: 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 
$\lim_{n\to\infty}(1-\frac{\ln n}{n})^n=\lim_{n\to\infty}((1-\frac{\ln n}{n})^{\frac{n}{\ln n}})^{\ln n}=0$.
A: Hint: consider
$$
\log \left( \left( 1-\frac{\log n}{n} \right)^n \right) = n \log \left( 1-\frac{\log n}{n} \right)
$$
and prove (or recall) that
$$
\lim_{n \to +\infty} \frac{\log n}{n} =0.
$$
Since $\log (1+\varepsilon ) \approx \varepsilon$ as $\varepsilon \to 0$, ...
A: Let $A=\lim_{n\to\infty}(1-\ln n/n)^n$. Take $\ln$  both sides,we get $lnA=\lim_{n\to\infty}n ln(1-\ln n/n)$.Now put $x=1/n$,it will give you $lnA=lim_{y\to0^+}\frac{ln(1+y\ln y)}{y}$.Using L'Hopital's Rule,you will get this limit=$-\infty$.Therefore,$\ln A=-\infty \implies A=0$.So, the required limit is $0$.
A: Your approach is wrong (in particular, how could your result contain $n$?).
Notice that $(1-\frac{\ln n}{n})^n=e^{n\ln(1-\frac{\ln n}{n})}$ and $\frac{\ln n}{n}\rightarrow 0$, so $n\ln(1-\frac{\ln n}{n})\sim-\ln n\rightarrow -\infty$. 
Moreover $\displaystyle\lim_{n\to-\infty}e^n=0$.
Conclusion : $\displaystyle\lim_{n\to+\infty}(1-\frac{\ln n}{n})^n=0$.
