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?

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
$$\color{blue}{\frac{1}{2}} = \frac{1}{n}\int_0^1 nx \; dx \leq \frac{1}{n}\int_0^1 \ln (1+ e^{nx}) \; dx \leq \frac{1}{n}\int_0^1 nx \; dx  + \frac{1}{n}\int_0^1 \ln 2 \; dx =\color{blue}{\frac{1}{2}} + \frac{\ln 2}{n}$$
