Evaluating $\lim_{n \to \infty} \left ( \sqrt[n]{n} + \frac 1 n \right )^{\frac n {\ln n}}$ I have trouble evaluating this limit:
$$\lim_{n \to \infty} \left ( \sqrt[n]{n} + \frac 1 n \right )^{\frac n {\ln n}}$$
I cannot use series expansion... I also tried to rewrite it as $\large e^{\ln (\cdots)}$, without getting anywhere. It's the $\sqrt[n]{n}$ factor that troubles me the most. I can't get rid of it.
 A: Factor out $n^{1/n}$ via $\log(n^{1/n}+n)=\log(n^{1/n})+\log(1+1/n^{1+1/n})$ and simplify. You can taylor expand the second $\log$ if necessary. It looks the answer will be $e$. 
A: 
With no trick, using only that $\mathrm e^x=1+x+o(x)$ and $\log(1+x)=x+o(x)$ when $x\to0$.

(Note that these two expansions are actually equivalent since $\exp$ and $\log$ are mutually inverse.)


*

*$\sqrt[n]{n}=\mathrm e^{\log n/n}=1+\log n/n+o(\log n/n)$

*$1/n=o(\log n/n)$

*$\sqrt[n]{n}+1/n=1+\log n/n+o(\log n/n)$

*$\log(\sqrt[n]{n}+1/n)=\log n/n+o(\log n/n)$

*$(n/\log n)\log(\sqrt[n]{n}+1/n)=1+o(1)$

*$\exp(1+o(1))\to\mathrm e$

A: Use the fact that if $a_n \rightarrow a$ then 
$$\left(1+\frac{a_n}{n}\right)^n \rightarrow e^a$$
We write our expression as
$$\left(\sqrt[n]{n}+\frac{1}{n}\right)^{\frac{n}{\ln n}}=\left(1+\frac{n(\sqrt[n]{n}-1)+1}{n}\right)^{\frac{n}{\ln n}}=
\left(1+\frac{a_n}{n/\ln n}\right)^{\frac{n}{\ln n}}$$
where 
$$a_n=\frac{n(\sqrt[n]{n}-1)+1}{\ln n}$$ so it is reduced to finding this last limit.
$\frac{1}{\ln n}\rightarrow 0$ so we want the limit of $\frac{n(\sqrt[n]{n}-1)}{\ln n}$
But this follows immediately from 
$$\lim\limits_{x\rightarrow 0} \frac{e^x-1}{x}=1$$ let $x=\frac{\ln n}{n}$
 to get
$$\lim\limits_{n\rightarrow \infty } \frac{\sqrt[n]{n}-1}{\ln n/n}=1$$ therefore 
$a_n \rightarrow 1$ and our limit for the original expression will be $e$. In agreement with the other answers.
