Show that $\int_{-\infty}^{\infty}(1+\frac{x^2}{n})^{-n}dx\rightarrow \int_{-\infty}^{\infty} e^{-x^2}dx$ 
Show that $\int_{-\infty}^{\infty}(1+\frac{x^2}{n})^{-n}dx\rightarrow
\int_{-\infty}^{\infty} e^{-x^2}dx$

I thought the best way to go about this is to show uniform convergence of $f_n(x)=(1+\frac{x^2}{n})^{-n}$ to $f(x)=e^{-x^2}$, which would justify swapping the limit and integral (I think? This would hold on a bounded interval, not so sure about this). Dini's theorem would work had it been a closed interval, but since we're after uniform convergence in all of $\Bbb R$ I can't seem to find a way to work this out.
 A: As discussed in the comments you can apply the dominated convergence theorem to prove the convergence of the integrals. Here's an alternative solution using uniform convergence:
Let $f(x) = \mathrm{e}^{-x^2} \, , \,  x \in \mathbb{R}$ . For $R \geq 1$ and $n \in \mathbb{N}$ we have
\begin{align}
\left|~ \int \limits_{\mathbb{R}\setminus [-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| &\leq 2 \int \limits_R^\infty \left[\mathrm{e}^{-x^2} + \frac{1}{\left(1+\frac{x^2}{n}\right)^n}\right] \, \mathrm{d} x \\
&\leq 2 \int \limits_R^\infty \left[\mathrm{e}^{-x} + \frac{1}{x^2}\right] \, \mathrm{d} x \\
&= 2 \left(\mathrm{e}^{-R}+\frac{1}{R}\right) \, .
\end{align} 
Now let $\varepsilon > 0$. Choose $R \geq 1$ such that $2(\mathrm{e}^{-R} + \frac{1}{R}) < \frac{\varepsilon}{2}$ . Since $[-R,R]$ is compact, we can use Dini's theorem to conclude that $f_n \rightarrow f$ uniformly on $[-R,R]$ as $n \rightarrow \infty$ . But then the sequence of integrals over this interval converges as well, so we can find an $N \in \mathbb{N}$ such that
$$ \left|~ \int \limits_{[-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| < \frac{\varepsilon}{2} $$
holds for every $n \geq N$ . This implies
\begin{align} \left|~ \int \limits_{\mathbb{R}} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| &\leq \left|~ \int \limits_{\mathbb{R}\setminus [-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| + \left|~ \int \limits_{[-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| \\
&< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
\end{align}
for $n \geq N$ as was to be shown.  
