Find $\lim \mathbb{E} \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$ Consider $Z_{n} = \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$, in which $X_n$ denotes a Gaussian random variable with zero expected value and variance equals $1/n$. Now we want find 
$\lim_{n \to \infty} \mathbb{E} \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$.
I thought about find characteristic function of $Z_{n}$ and so it will be easy to find expectation. But I guess there should be easier approach. Any ideas ?
 A: I think it could be cracked by real analysis method. For convenience note $X_n$ by $f_n$. Then $EX_n^2=\int f_n^2dP=1/n$, and $f_n\overset{P}{\to}0$.
For $\delta>0$, set $$M(\delta)=\sup_{|x|\leq\delta}\frac{x^2}{\log(1+x^2)}$$ $$m(\delta)=\inf_{|x|\leq\delta}\frac{x^2}{\log(1+x^2)}$$
Note that $$\lim_{\delta\to0}M(\delta)=\lim_{\delta\to0}m(\delta)=1$$
We estimate this formula separately
\begin{eqnarray}
\int\frac{f_n^2}{\log(1+f_n^2)}dP&=&\int_{\{|f_n|\leq\delta\}}\frac{f_n^2}{\log(1+f_n^2)}dP+\int_{\{|f_n|>\delta\}}\frac{f_n^2}{\log(1+f_n^2)}dP
\end{eqnarray}
Fix $\delta$, as $n\to\infty$, for the upper bound
$$\leq P\{|f_n|\leq\delta\}M(\delta)+\frac{1}{\log(1+\delta^2)}\frac{1}{n}
\to M(\delta)$$
for the lower bound
$$\geq P\{|f_n|\leq\delta\}m(\delta)
\to m(\delta)$$
Let $\delta\to 0$, so we proved
$$\lim_{n\to\infty}\mathbb{E}\frac{X_n^2}{\log(1+X_n^2)}=1.$$
By the way, actually we don't need $X_n$ to be Gaussian r.v.
A: The function $f(x) = \frac{x}{\log(1+x)}  = \int_{0}^{1} (1+x)^s \, ds $ is concave and increasing on $[0,\infty)$. So
$$ f(0) \leq \mathbf{E}\left[ \frac{X_n^2}{\log(1+X_n^2)} \right] = \mathbf{E}[f(X_n^2)] \stackrel{\text{(Jensen)}}{\leq} f(\mathbf{E}[X_n^2]). $$
The conclusion follows from the squeezing theorem by noticing that $\mathbf{E}[X_n^2] = \frac{1}{n} \to 0$ as $n\to\infty$.
A: There's a simpler proof that the limit is $1$ if you use Gaussianity of the $X_n$.
Wlog you may as well assume that $X_n = n^{-1/2}Y$, where $Y \sim N(0,1)$, since this does not change the expectation.
Now, since $\lim_{u \to 0} \frac{u}{\log(1+u)}=1$, it follows that $Z_n = \frac{n^{-1}Y^2}{\log(1+n^{-1}Y^2)} \to 1$ a.s.
Now, if we can show that $\sup_n Z_n$ is bounded above by an integrable r.v., then we are done since we can just apply DCT. To prove this, just note that $\frac{u}{\log(1+u)} \leq 1+u$, so that $Z_n \leq 1+n^{-1}Y^2 \leq 1+Y^2$.
