Finding $\lim_{n \to \infty}(n^{1\over n}+{1\over n})^{n\over \ln n}$ 
$$\lim_{n \to \infty}\left(n^{1\over n}+{1\over n}\right)^{n\over \ln n}$$

The limit is same as
$$e^{\displaystyle{\lim_{n \to \infty}}{n^{(1+{1\over n})}-n+1\over \ln n}}$$
But I am stuck here , I noticed that if I take $n$ common from the numerator then it is of the form $0 \times \infty$ , but I can't seem to solve further , using L'hopital's rule is too hectic or perhaps not meant to solve this one . I couldn't use any of the Maclaurin expansions either in this case .
Could someone please guide me in this ?
Thanks !
 A: Here is simple approach  for this limit.
$$ \lim_{n\to \infty} \left(\sqrt[n]{n} +\frac{1}{n} \right)^{\frac{n}{\ln n}} = \lim_{n\to\infty} \left[ n^{\frac{1}{\ln n}}\left(1+\frac{1}{n \sqrt[n]{n}}\right)^{\frac{n}{\ln n}}\right]=\lim_{n\to\infty}  n^{\frac{1}{\ln n}} \exp\left(\lim_{n\to\infty} \frac{1}{\sqrt[n]{n} \ln n}\right) =e \times 1 = e $$
Since $$ \lim_{n\to\infty} n^{\frac{1}{\ln n}} = \exp\left(\lim_{n\to \infty} \frac{\ln  n}{\ln n}\right)=e$$
In the latter limit we use the fact that $\displaystyle \sqrt[n]{n} =1$ so as $n\to \infty $  and $\ln n\to \infty$ which tell us that $\frac{1}{\sqrt[n]{n} \ln  n}$  is decreasing sequence for all $ n>1$  and it's limit  around  large $ n$   is $\displaystyle  \lim_{n\to\infty}\frac{1}{\ln n} =0$ .
A: $$a_n=\left(n^{1\over n}+{1\over n}\right)^{n\over \log( n)}\implies \log(a_n)={n\over \log( n)}\log\left(n^{1\over n}+{1\over n}\right)$$ By Taylor
$$\log\left(n^{1\over n}+{1\over n}\right)=\frac{\log (n)+1}{n}-\frac{2\log
   (n)+1}{2n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(a_n)=1+\frac 1 {\log(n)}-\frac{2\log
   (n)+1}{2n\log(n)}+O\left(\frac{1}{n^2}\right)$$
$$a_n=e^{\log(a_n)}=e \, \exp\left(\frac 1 {\log(n)}-\frac{2\log
   (n)+1}{2n\log(n)}+O\left(\frac{1}{n^2}\right) \right)$$ xhish shows the limit and how it is approached.
Moreover, this gives you a shortcut method for an estimate of $a_n$. Suppose $n=10$; the truncated expression would give $3.716$ while the exact value is $3.788$.
It would give a relative error lower than $1$% for $n \geq 15$ and lower than $0.1$% for $n \geq 54$.
A: Your first step is correct, as in taking the exponential and the logarithm.
However, the next quantity is not correct. Take $L$ to be the limit in question, then
$$L=\exp\left(\lim_{n\to\infty}\frac{n}{\log n}\log\left(n^{\frac1n}+\frac1n\right)\right).$$
I can't quite piece how did you get to your quantity but that's beside the point. Now rewrite
$$\frac1n=n^{-1+\left(\frac1n-\frac1n\right)}=n^{\frac1n}\cdot n^{-\left(1+\frac1n\right)}.$$
Using the product rule for logarithms, the $\log$ right there becomes
$$\log\left(n^{\frac1n}+\frac1n\right)=\log\left(n^{\frac1n}+n^{\frac1n}\cdot n^{-\left(1+\frac1n\right)}\right)=\frac{\log n}n+\log\left(1+n^{-\left(1+\frac1n\right)}\right).$$
Back in the original expression we substitute back to get
\begin{align*}
L&=\exp\left(\lim_{n\to\infty}\frac{n}{\log n}\left(\frac{\log n}n+\log\left(1+n^{-\left(1+\frac1n\right)}\right)\right)\right)\\
&=\exp\left(1+\lim_{n\to\infty}\frac{n}{\log n}\cdot\log\left(1+n^{-\left(1+\frac1n\right)}\right)\right).
\end{align*}
We now treat the last logarithm, so as an strategy, we know that $\frac{n}{\log n}$ goes to $\infty$ as $n$ grows since $n\gg\log n$. Therefore its inverse goes to zero, then we would like the other quantity to go to zero as well to apply L'Hôpital's rule. It is enough to show that $n^{-\left(1+\frac1n\right)}$ tends to zero as $n$ grows.
$$\lim_{n\to\infty}n^{-\left(1+\frac1n\right)}=\exp-\left(\lim_{n\to\infty}\left(1+\frac1n\right)\log(n)\right)=\exp-((1+0)\infty)=0.$$
However, I feel that trying to apply L'Hôpital's rule here would be too convoluted. If anyone is willing to expand on this procedure I would be grateful.
I'll be back once I see how to get around this.
