Let $g$ be a bounded measurable function on $[a, b]$. Then compute the limit $$\lim_{n \to \infty} \int_a^b \frac{g(x)e^{nx}}{1+e^{nx}} dx$$
I write the given integral as $$\lim_{n \to \infty} \int_a^b \frac{g(x)e^{nx}}{1+e^{nx}} dx = \int_a^b g(x) dx - \lim_{n \to \infty} \int_a^b \frac{g(x)}{1+e^{nx}} dx$$ and i guess $$\lim_{n \to \infty} \int_a^b \frac{g(x)}{1+e^{nx}} dx = 0$$ Am i correct? How to prove the guess?
Yes, you can use Lebesgue dominated convergence here since we can dominate the function inside the integral. Since $g(x)$ is bounded, there exists a constant $c$ s.t. $|g(x)|<c$. We also have that $1+e^{nx} > 1$, so we can combine these to get that the function inside the integral is dominated by a constant function, $$\frac{|g(x)|}{1+e^{nx}} < c$$ This allows us to swap the limit and the integral, so now it comes down to evaluating $$\lim_{n\to \infty} \left(\frac{1}{1+e^{nx}}\right) = \begin{cases} 0 & nx>0\\ \frac{1}{2} & nx=0\\ 1 & nx<0 \end{cases}$$ These cases are all fairly straightforward, though you'll have to be careful when the sign of $nx$ swaps, so when $a$ is negative and $b$ is positive for example.