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.
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.