Question involving limit How to find the following limit:
$$\lim_\limits{n\to\infty} (ne\sqrt[n]{\ln{(1+e^n)}-n}-n)$$
Thanks in advance!
 A: Hint:
$$\ln(1+e^n)-n=n+\ln(e^{-n}+1)-n=e^{-n}-\frac12e^{-2n}+\frac13e^{-3n}-\cdots=e^{-n}\left(1-\frac{e^{-n}}2+\frac{e^{-2n}}3-\cdots\right).$$
Then
$$ne\sqrt[n]{\ln(1+e^n)-n}-n=nee^{-1}\sqrt[n]{1-\frac{e^{-n}}2+\frac{e^{-2n}}3-\cdots}-n\\
=n\left(1-\frac{e^{-n}}{2n}+\frac{e^{-2n}}{3n}-\cdots\right)-n.$$
A: I will start from a slightly modified limit:
$$\lim_{n\to\infty}\left(e^n \left(n-ne\sqrt[n]{\ln{(1+e^n)}-n}\right) \right)=\frac12
$$
because that will immediately say that 
$$
\lim_{n\to\infty}\left( ne\sqrt[n]{\ln{(1+e^n)}-n}-n \right)=0
$$
So how do we show this?
Start from 
$$
\log(1+e^n) = \log\left( (1+e^{-n} )e^n\right)=\log(1+e^{-n}) + \log(e^n)  =n+\log(1+e^{-n})
$$
Then 
$$
\ln{(1+e^n)}-n= \log(1+e^{-n})
$$
Now use the expansion for small arguments of $\log(1+u)$ and the binomial theorem (both valid becasue $e^{-n}$ will be tiny) to expand 
$(\log(1+e^{-n}))^{1/n}$
to first order:
$$
(\log(1+e^{-n}))^{1/n} \approx \left( e^{-n}-\frac{e^{-2n}}{2} \right)^{1/n}
= e^{-1} \left(1-\frac{e^{-n}}{2}\right)^{1/n} \approx e^{-1} \left( 
1-\frac{1}{n}\frac{e^{-n}}{2} \right) 
$$
So 
$$ ne
\left( (\log(1+e^{-n}))^{1/n} \right) \approx n-\frac{e^{-n}}{2} + O(e^{-2n})
$$ 
and from that follows 
$$\lim_{n\to\infty}\left(e^n \left(n-ne\sqrt[n]{\ln{(1+e^n)}-n}\right) \right)=\frac12
$$
A: The limit is same even if $n$ is a real variable rather than a positive integral variable. And then we switch to symbol $x$ instead of $n$. We have
\begin{align}
L &= \lim_{n \to \infty}\left(ne\sqrt[n]{\log(1 + e^{n}) - n} - n\right)\notag\\
&= \lim_{x \to \infty}\left(ex\{\log(1 + e^{x}) - x\}^{1/x} - x\right)\notag\\
&= \lim_{x \to \infty}\left(ex\{\log(1 + e^{-x})\}^{1/x} - x\right)\notag\\
&= \lim_{x \to \infty}x\left(\exp\left(\frac{\log\log(1 + e^{-x}) + x}{x}\right) - 1\right)\notag\\
&= \lim_{x \to \infty}x\left(\exp\left(\frac{\log\{e^{x}\log(1 + e^{-x})\}}{x}\right) - 1\right)\notag\\
&= \lim_{x \to \infty}x\cdot\dfrac{\exp\left(\dfrac{\log\{e^{x}\log(1 + e^{-x})\}}{x}\right) - 1}{\dfrac{\log\{e^{x}\log(1 + e^{-x})\}}{x}}\cdot \dfrac{\log\{e^{x}\log(1 + e^{-x})\}}{x}\notag\\
&= \lim_{x \to \infty}x\cdot 1\cdot \dfrac{\log\{e^{x}\log(1 + e^{-x})\}}{x}\notag\\
&= \lim_{x \to \infty}\log\{e^{x}\log(1 + e^{-x})\}\notag\\
&= 0\notag
\end{align}
Note that when $x \to \infty$ then $t = e^{-x} \to 0$ and hence $e^{x}\log(1 + e^{-x}) = \dfrac{\log(1 + t)}{t} \to 1$ and therefore $$\log\{e^{x}\log(1 + e^{-x})\} \to 0$$ and we have used this fact in the above evaluation. As usual most complicated looking limit problems often do not necessitate the use of powerful methods like Taylor's series and L'Hospital's Rule and simpler tools like standard limits suffice.
