Solve $\lim_{n\rightarrow 0}\frac{1}{n}\int_{0}^{1}\ln(1+e^{nx})dx$ $$\lim_{n\rightarrow 0}\frac{1}{n}\int_{0}^{1}\ln(1+e^{nx})dx$$
My try:
$$\frac{b-a}b\leq \ln b-\ln a\leq \frac{b-a}a \implies \frac{1}{1+e^{nx}}\leq \ln(1+e^{nx})-\ln e^{nx}\leq \frac1{e^{nx}}$$
Then I integrated and multiplied by $\frac{1}{n}$ and I got:
$$\frac{1}{n}\int _0^1\frac{1}{1+e^{nx}}dx\leq \frac{1}{n}\int _0^1[\ln(1+e^{nx})-\ln e^{nx}]dx\leq \frac{1}{n}\int _0^1 \frac{1}{e^{nx}}dx$$
How to continue? There is an easier method to solve this limit?
 A: Hint: The two functions $x\mapsto e^{-nx}$ and $x\mapsto \frac{1}{1+e^{nx}}$ are bounded over $[0,1]$

Your last inequality can be rewritten as:
$${m\over n}\le \bigg({1\over n}\int_{0}^{1}{\log(1+e^{nx})dx}\bigg)-{1\over 2}\le {M\over n}$$ where
$$m=\inf_{x\in[0,1]}\big({\frac{1}{1+e^{nx}}}\big)={1\over{1+e^n}}$$ and 
$$M=\sup_{x\in[0,1]}\big({\frac{1}{e^{nx}}}\big)=1.$$ From here you can use the celebrated squeeze theorem.
A: You may squeeze the integral as follows using


*

*$nx = \ln e^{nx} \leq \ln (1+ e^{nx})$ and

*$\ln (1+ e^{nx}) = nx + \ln \left(1 +\frac{1}{e^{nx}}\right)\stackrel{0\leq x \leq 1}{\leq} nx + \ln 2$
So, you get
$$\color{blue}{\frac{1}{2}} = \frac{1}{n}\int_0^1 nx \; dx \leq \frac{1}{n}\int_0^1 \ln (1+ e^{nx}) \; dx \leq \frac{1}{n}\int_0^1 nx \; dx  + \frac{1}{n}\int_0^1 \ln 2 \; dx =\color{blue}{\frac{1}{2}} + \frac{\ln 2}{n}$$
A: By integration by parts,
\begin{eqnarray}
\frac{1}{n}\int_{0}^{1}\ln(1+e^{nx})dx&=&\frac1nx\ln(1+e^{nx})\bigg|_0^1-\int_0^1\frac{xe^{nx}}{1+e^{nx}}\\
&=&\frac1nx\ln(1+e^{nx})\bigg|_0^1-\int_0^1\frac{x(1+e^{nx})-x}{1+e^{nx}}\\
&=&\frac1n\ln(1+e^n)-\int_0^1xdx+\int_0^1\frac{x}{1+e^{nx}}dx\\
&=&\frac1n\ln(1+e^n)-\frac12+\int_0^1\frac{x}{1+e^{nx}}dx.
\end{eqnarray}
Noting
$$ \lim_{n\to\infty}\frac1n\ln(1+e^n)=1$$
and
$$ 0<\int_0^1\frac{x}{1+e^{nx}}dx\le\int_0^1e^{-nx}dx=\frac{1-e^{-n}}{n}\to 0 \text{ as }n\to\infty$$
one has
$$\lim_{n\rightarrow 0}\frac{1}{n}\int_{0}^{1}\ln(1+e^{nx})dx=\frac12. $$
