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 ? 
 A: 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
$$ \lim_{n\to\infty}\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx=\lim_{n\to\infty}\ln^{3} (2+\frac{1}{\xi^{2}(n)})\ln(\frac{1+2n}{1+n})=\ln^42.$$
A: 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.
A: You can try to squeeze the integrand suitably. The fraction $n/(1+nx)$ lies between $1/x-1/(nx^2)$ and $1/x$ and $\log^3(2+1/(nx)^2)$ lies between $\log^32$ and $(\log 2+1/(2nx)^2)^3$. Thus the integrand lies between $$\log^32\left(\frac{1}{x}-\frac{1}{nx^2}\right)$$ and $$\frac{1}{x}\left(\log 2+\frac{1}{(2nx)^2}\right)^3$$ Both the above expressions can be written as $\dfrac{\log ^32}{x}$ and a finite number of terms of the form $\dfrac{k} {n^ax^b} $ where $k$ is a constant and $a, b$ are positive integers. Clearly the integrals of such terms over interval $[1,2]$ tend to $0$ because of factor $1/n^a$ and therefore the desired limit is $\log^42$ as expected. 
