Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $ 
Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $
where $f:[-1,1] \to \mathbb{R}$ is a continuous function.

I have no clue how to do it... Any hints?

Following the hints, by letting $t=\sqrt{n}x$, we have that
$$\sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ dx=\int\limits_{\mathbb{R}} e^{-\frac{t^{2}}{2}}f\left(\frac{t}{\sqrt{n}}\right)I_{[-\sqrt{n},\sqrt{n}]}(t)\ dt$$
Now for all $t\in\mathbb{R}$,
$$F_n(t):=e^{-\frac{t^{2}}{2}}f\left(\frac{t}{\sqrt{n}}\right)I_{[-\sqrt{n},\sqrt{n}]}(t)\to F(t):=e^{-\frac{t^{2}}{2}}f(0)$$
and 
$$|F_n(t)|\leq Me^{-\frac{t^{2}}{2}}$$
where $M=\sup_{x\in[-1,1]}|f(x)|$. 
Since $Me^{-\frac{t^{2}}{2}}$ is integrable over $\mathbb{R}$, by the Dominated Convergence Theorem
$$\int_{\mathbb{R}}F_n(t) \ dt\to \int_{\mathbb{R}}F(t) \ dt=f(0)\int_{\mathbb{R}}e^{-\frac{t^{2}}{2}}\ dt=f(0)\cdot \sqrt{2\pi}.$$
