$\lim_{\varepsilon\to 0}\int_{-1}^1 \frac 1 {\sqrt{2\pi \varepsilon}} e^{-\frac{x^2}{2\varepsilon}}\ell(x) \, dx={}$??? What is the limit of the following:
$$\lim_{\varepsilon\to 0}\int_{-1}^1 \frac{1}{\sqrt{2\pi \varepsilon}} e^{-\frac{x^2}{2\varepsilon}} \ell(x) \,dx.$$
where $\ell$ is a bounded and nice function ($\ell\in C^\infty$)? Attention the problem is in the neighborhood of $0$, if we are not in the neighborhood of $0$ the limit is easy.
 A: Let $u = x/\sqrt\varepsilon,$ so that $du = dx/\sqrt\varepsilon$ and as $x$ goes from $-1$ to $1$ then $u$ goes from $-1/\sqrt\varepsilon$ to $1/\sqrt\varepsilon.$ Now recall that
$$
\lim_{\varepsilon\,\downarrow\,0} \int_{-1/\sqrt\varepsilon}^{+1/\sqrt\varepsilon} \frac 1 {\sqrt{2\pi}} e^{-u^2/2} \, du = 1.
$$
The given integral is equal to
$$
\int_{-1/\sqrt\varepsilon}^{+1/\sqrt\varepsilon} \frac 1 {\sqrt{2\pi}} e^{-u^2/2} \ell(u\sqrt\varepsilon)\, du.
$$
It was given that $\ell$ is bounded, so for some $M>0$ we have
$$
\frac 1 {\sqrt{2\pi}} e^{-u^2/2} \ell(u\sqrt\varepsilon) \mathbf 1_{|u|\,\le\,1/\sqrt\varepsilon} \le \frac 1 {\sqrt{2\pi}} e^{-u^2/2}\cdot M, \tag 1
$$
and this is a non-negative function whose integral is finite. Since the functions on the left side of $(1),$ for $\varepsilon>0,$ are all bounded by this one function not depending on $\varepsilon$ whose integral is finite, the limit of the integral of the left side over $\mathbb R$ is equal to the integral of the limit, i.e. it is equal to
$$
\int_{\mathbb R} \frac 1 {\sqrt{2\pi}} e^{-u^2/2} \ell(0)\, dx = \ell(0).
$$
A: First: $\lim_{\epsilon\to 0}\int_{-1}^1\frac{1}{\sqrt{2\pi \epsilon}}e^{-\frac{x^2}{2\epsilon}}dx=1$.    Also, as $\epsilon \to 0$, the integrand $\to 0$ for $|x| \gt \delta$ for any $\delta \gt 0$. 
Assume $l(x)$ is bounded and also continuous at $x=0$. So using continuity of $l(x)$, we can substitute $l(0)$ for $l(x)$ in the integrand.  Net result, the limit of the integral$=l(0)$. 
