How to show $\int_{\mathbb{R}^d}  u(x) e^{-|x|^2}  (e^{-|x|^2 / n} - 1)^2 dx \rightarrow 0$, if $u\in L^2(\Bbb R^n)$? How can I prove that 
$$
\int_{\mathbb{R}^d} u(x) e^{-|x|^2} (e^{-|x|^2 / n} - 1)^2 dx \rightarrow 0
$$
as $n \rightarrow \infty$?  Here $u \in L^2(\mathbb{R}^n)$.
I'm thinking the dominated convergence theorem, but I don't know how to bound the integrand.
Thanks.
 A: Added latter:(a simpler proof) We have 
$$\left(\exp\left(-\frac{|x|^2}n\right)\right)^2=\left(\int_0^{-\frac{|x|^2}n}e^tdt\right)^2\leq \frac{|x|^4}{n^2},$$
hence 
$$\left|\int_{\mathbb R^d}u(x)f_n(x)dx\right|\leq \frac 1{n^2}||u||_{L^2}\left(\int_{\mathbb R^d}e^{-2|x|^2}|x|^8dx\right)^{\frac 12},$$
and we can conclude since $e^{-2|x|^2}|x|^8$ is integrable. It's shows that the sequence $\{f_n\}$ converges strongly to $0$.

We can use dominated or the fact that the functions with compact support are dense in $L^2$. Put $f_n(x):=e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2$. Since $\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2\leq 1$ for all $x$ and for all $n$, we have $\lVert f_n\rVert_{L^2}\leq \sqrt{\int_{\mathbb R^d}e^{-2|x|^2}dx}=:M$. Let $u\in L^2(\mathbb R^d)$ and $\varepsilon >0$. Let $g$ continuous with compact support $K$ such that $\|u-g\|_{L^2}\leq\frac{\varepsilon}{M}$ (consequence of the monotone convergence theorem). Then 
\begin{align*}
\left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|&\leq \int_{\mathbb R^d}|u-g|f_n(x)dx +\int_{\mathbb R^d}|g|f_n(x)\\
&\leq \|u-g\|_{L^2} M+\sup |g| \int_Ke^{-|x|^2}\frac{\sup_{w\in K} |w|^4}{n^2}dx\\
&\leq \varepsilon+\sup |g|\sup_{w\in K} |w|^4\frac 1{n^2},
\end{align*}
so for all $\varepsilon >0$:
$$\limsup_n\left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|\leq \varepsilon,$$
hence 
$$\lim_{n\to\infty}\left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|=0.$$
