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 ? 

Using the inequality $\ln (1+x) \leq x$ for $x>0$ we get the bound $n\frac {(1+\frac 1 {n^{2}})^{3}} {1+n}$ for the integrand. Since this quantity is bounded (by $8$, for example,) we can apply Bounded Convergence Theorem.
