Finding the asymptotes of an integral I need to find the asymptote of $$\int\limits_{0}^{\infty} \frac{\ln(1 + \frac{x}{\sqrt{n}})}{x + x^3} dx$$
I've taken $b_{n} = n^{-1/2}$ which reduces the problem to finding,
$$\lim_{n\to\infty}\ \sqrt{n}\int\limits_{0}^{\infty} \frac{\ln(1 + \frac{x}{\sqrt{n}})}{x + x^3} dx$$ but now I'm stuck. Can I bring the limit inside the integral? Am I supposed to show that the limit is equal to 1 so that $\lim_{n\to\infty} \frac{a_n}{b_n} = 1$?

I would use squeezing, with the inequalities
$$
\frac{x}{x+1}<\ln(1+x)<x.
$$
Inserting and calculating, you will find that
$$
\frac{\sqrt{n}\pi-\ln n}{2+2n}\leq \int_0^{+\infty}\frac{\ln(1+x/\sqrt{n})}{x+x^3}\,dx\leq \frac{1}{\sqrt{n}}\frac{\pi}{2}.
$$
It follows that
$$
\lim_{n\to+\infty}\sqrt{n}\int_0^{+\infty}\frac{\ln(1+x/\sqrt{n})}{x+x^3}\,dx=\frac{\pi}{2}.
$$
