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 ?

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 $n\to\infty$.
