Evaluate : $\lim\limits_{n\to +\infty}\int\limits_n^{2n}\frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx$ Problem : 
Evaluate : 
$$\lim\limits_{n\to +\infty}\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx$$
My attempt : 
$$y=\frac{x}{n}$$ 
Then :
$$I(n)=\int\limits_1^2 n\frac{\ln^{3}(2+\frac{1}{(ny)^{2}})}{1+nx}dx$$
So : 
$$\lim\limits_{n\to +\infty}I(n)=\int_1^2 \frac{\ln^{3}(2)}{x}dx$$
$$=\ln^{4}(2)$$ 
But my question I can take limits inside the integral ? 

By the Mean Value Theorem for integrals, one has
$$ \int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx=\ln^{3} (2+\frac{1}{\xi^{2}(n)})\int\limits_n^{2n} \frac{1}{1+x}dx=\ln^{3} (2+\frac{1}{\xi^{2}(n)})\ln(\frac{1+2n}{1+n})$$
for some $\xi(n)\in(n,2n)$. Noting that, as $n\to\infty$, $\xi(n)\to\infty$, one has
$$ \lim_{n\to\infty}\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx=\lim_{n\to\infty}\ln^{3} (2+\frac{1}{\xi^{2}(n)})\ln(\frac{1+2n}{1+n})=\ln^42.$$
