Limit of a sequence of integrals Determine
$$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n}$$
Since $e^{x} \geq t+1$ we know that $e^{x^2}-1 \geq 2t +t^2 \geq t^2$ if $t\geq 0$
So due to symmetry, we should be able to use DCT?
But my solution is $$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n} = \int_{-\infty}^\infty \lim _{n \rightarrow \infty} \frac{dx}{n(e^{x^2}-1) +1/n} = 0$$. Which is the wrong answere.
Can someone help me out here?
We have not startet with variable change in measure theory, so please do not use that.

We start from the the inequality $ e^y \le 1+y +\frac {3y^2} {2}$ if $ y \ge 0,\, y \le 1$. This follows from the Taylor's theorem 
  in the Lagrange form. Next, we obtain with help of Maple
  $$J_n:=\int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \ge \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2+3nx^4/2 +1/n}=$$ $$ \frac {1}{\sqrt {{n}^{2}-6}\sqrt { \left( \sqrt {{n}^{2}-6}-n
 \right) n}\sqrt { \left( \sqrt {{n}^{2}-6}+n \right) n}}$$ $$-2\,n\sqrt {3} \left( \mathop{\rm arctanh} \left( {\frac {\sqrt {3n}}{\sqrt { \left( \sqrt {{n}^{2}-6}-n \right) n}}} \right) \sqrt {
 \left( \sqrt {{n}^{2}-6}+n \right) n}+$$
 $$ \left.\sqrt { \left( \sqrt{{n}^{2}-6
}-n \right) n}\arctan \left( {\frac {\sqrt {n}\sqrt {3}}{\sqrt {
 \left( \sqrt {{n}^{2}-6}+n \right) n}}} \right)  \right)\to \pi,\, n \to \infty.(1)
 $$
On the other hand,
  $$J_n= \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \le \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2 +1/n} =\frac 1 n \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{x^2 +1/n^2}=2\arctan(\sqrt{n}) \to \pi,\,n \to \infty .(2)$$ 
  It remains to estimate the tails: $$\int_{-\infty}^{-\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n}+ \int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{n(e^{x^2}-1) +1/n}\le 2\int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{nx^2} =\frac 2 {\sqrt{n}} \to 0,\, n \to \infty .(3)$$  Combining (1)-(3), we deduce that the limit under consideration equals $\pi$.
