LEt $f:\Omega\to[0,\infty]$ measurable s.t $0<c:=\int_{\Omega}fd\mu<\infty$ . Prove that $$\lim_{n\to\infty}\int_{\Omega}n\log\bigg(1+\big(\frac f n\big)^a\bigg)d\mu=\cases{c\quad a=1\\\infty\quad 0<a<1\\0\quad 1<a<\infty}$$
About the first case, I got upper bound by $$\lim\int n\log(1+\frac f n)\le\lim\int n\log(\exp(\frac f n))=\int f=c$$ but on the other side I can get that $$\lim_{n\to\infty}\int_{\Omega}n\log\bigg(1+\big(\frac f n\big)^a\bigg)d\mu\ge\lim\int n\log(\frac f n)$$ and by L'hspital rule to get the limit on the RHS is equal to $\int f$ but on the other hand, I'm not sure I can apply it here.
Aout the other cases I'm clueless. I need a reasining for switiching the limit and the integral, i.e. does $\lim_{n\to\infty}\int_{\Omega}n\log\bigg(1+\big(\frac f n\big)^a\bigg)d\mu\overset{?}{=}\int_{\Omega}\lim_{n\to\infty}n\log\bigg(1+\big(\frac f n\big)^a\bigg)d\mu$?
How can I solve it for the other cases? Can I switch the limit and the integral as described?
You can Switch the Limit and the integral. For computing the Limit of the Integrand, use L'Hospitals rule.
$\lim_{n \rightarrow \infty} \frac{log(1+(\frac{f}{n})^a)}{\frac{1}{n}} = \lim_{n \rightarrow \infty} \frac{(log(1+(\frac{f}{n})^a))'}{(\frac{1}{n})'}$.
For deriving the numerator by $n$: $(log(1+(\frac{f}{n})^a))'=\frac{-af^a/n^{a+1}}{1+(\frac{f}{n})^a}$.
The above Quotient can be simplified and results in: $\lim_{n \rightarrow \infty} \frac{af^a}{n^{a-1}+\frac{f^a}{n}}$.
Now consider the 3 different cases and take the Limit for every case.