Prove the convergence of the following integral: $$ \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx $$
I tried to use the obvious Weierstrass comparison test in the following way:
$$ \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx \leqslant \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2}dx = \lim\limits_{n\to\infty}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{nx^2}dx $$ However after thinking about a variable substitution I concluded that it would complicate the expression.
Question:
How should I prove the integral converges? What are your suggestions?
Edit: Tried a new approach but I would like someone to check it out. $\lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx=\lim\limits_{n\to\infty}\frac{2}{n}\int_\limits{0}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx$ since the function is even. Substitution $\sqrt{n}x=y$ So we get after some adjustments: $\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{1-e^{-y^2}}{y^2(1+y^2)}dy$ Then I am not sure about the following step: $1-e^{-y^2}\sim y^2$ So we would get $\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{{y^2}}{y^2(1+y^2)}dy=$$\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{1}{(1+y^2)}dy=0$
As in the comments indicated, We can just use the change of variables $x \sqrt{n} =y$ to get $$ \frac{1}{n} \int_{-n}^n \frac{1-\exp(-nx^2)}{x^2(1+nx^2)} =\frac{2}{\sqrt{n}} \int_0^{n^{3/2}} \frac{1-\exp(-x^2)}{x^2(1+x^2)} \mathrm{d} x. $$ Noting that $$0 \leq 1- \exp(-x^2) = \int_{0}^{x^2} \exp(-t) \, \mathrm{d} t \leq \min(x^2, \int_0^\infty \exp(-t) \mathrm{d} t) = \min(x^2,1)$$ we see that the last integral is bounded by $$\int_0^\infty \frac{\min(1,x^2)}{x^2(1+x^2)} \leq \int_0^1 \frac{1}{1+x^2} \mathrm{d} x+ \int_1^\infty \frac{1}{x^2(1+x^2)} < \infty.$$ Thus, we can conclude that the limes is $0$ and the convergence rate is $1/\sqrt{n}$.