How to compute the limit of the following integral?

Question

Given $b > 0$, let $g(x)$ be a continuous function defined on $[-b, b$]. What is the following limit?

$\lim_{N \rightarrow \infty} \frac{1}{\sqrt{N}} \int_{-b}^b e^{-\frac{Nx^{2}}{2}}g(x)\,dx $

We are told to make a change of variables to make things easier.

Answer 1

You can use the Lebesgue dominated convergence theorem. If $f_N(x) = e^{-Nx^2/2} g(x)$, since $g$ is bounded on $[-b,b]$, say by $M > 0$, then $|f_N(x)| \leq M e^{-Nx^2/2} \leq M e^{-x^2/2}$ for all $N$ so that the sequence is uniformly integrable on $[-b,b]$. Furthermore, $f_N(x) \to 0$ almost everywhere. Therefore, by the LDCT, $$ \begin{align*} \lim_{N \rightarrow \infty} \frac{1}{\sqrt{N}} \int_{-b}^b e^{-\frac{Nx^{2}}{2}}g(x)\,dx &= \left(\lim_{N \to \infty} \frac{1}{\sqrt{N}}\right)\left(\lim_{N\to\infty} \int_{-b}^b e^{-\frac{Nx^{2}}{2}}g(x)\,dx\right)\\ &= \left(\lim_{N \to \infty} \frac{1}{\sqrt{N}}\right)\left(\int_{-b}^b \lim_{N\to\infty} e^{-\frac{Nx^{2}}{2}}g(x)\,dx\right) \\ &= 0 \cdot 0 = 0. \end{align*} $$

Answer 2

Another approach is to notice that, the sequence $ \frac{e^{-\frac{N x^2}{2}}}{\sqrt{N}} $ converges uniformly to $0$ on the interval $[-b,b]$.