Distributional Limit $T_{g_{n}} \rightarrow \delta_{0}$ Consider $$
\psi(x)=\left\{\begin{array}{ll}
e^{-\frac{1}{1-x^{2}}}, & |x|<1 \\
0, & |x| \geqslant 1
\end{array}\right.
$$
and
$$
g_{n}(x)=\frac{\psi(n x)}{\displaystyle\int_{-\infty}^{\infty} \psi(n x) \: \mathrm{d} x}
$$
for $n \in \mathbb{N}$. I need to prove that $T_{g_{n}} \rightarrow \delta_{0}$ in $\mathscr{D}^{\prime}$ for $n \rightarrow+\infty$. ($T_{g_{n}}$ denotes the regular distribution induced by $g_n$).
I have tried to write down the action of $T_{g_n}$ on a test function $\phi(x)$ as:
$$\langle T_{g_n}, \phi(x)\rangle=\int_{-\frac 1n}^{\frac 1n} \frac{e^{-\frac{1}{1-n^2x^{2}}}}{\displaystyle\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2y^{2}}} \: \mathrm{d} y} \phi(x) \: \mathrm{d}x = \frac{1}{\displaystyle\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2y^{2}}} \: \mathrm{d} y} \int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} \phi(x) \: \mathrm{d} x $$
But I really don't know what to do from here. Any ideas?
 A: HINT:
Enforce the substitution $x\mapsto x/n$ to find that
$$\int_{-1/n}^{1/n}e^{-1/(1-n^2x^2)}\,\phi(x)\,dx=\frac1n \int_{-1}^1 e^{-1/(1-x^2)}\,\phi(x/n)\,dx$$
Now, divide by $\frac1n \int_{-1}^1 e^{-1/(1-x^2)}\,dx$ and apply the Dominated Convergence Theorem.


NOTE:
As mentioned by @paulgarrett in a comment, there is nothing particularly special about the Bump function $e^{-1/(1-x^2)}$ on $[-1,1]$.   For any non-negative function $\psi(x)\in C_C$ with $\psi(0)>0$, we see that for any $\phi\in C_C^\infty$
$$\lim_{n\to \infty}\frac{\int_{-\infty}^\infty \psi(nx)\phi(x)\,dx}{\int_{-\infty}^\infty \psi(nx)\,dx}=\phi(0)$$
A: Given $\varepsilon>0$, there is $\delta>0$ such that $x|<\delta$ implies  $|\phi(x)-\phi(0)|<\varepsilon$
$$\left|\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} (\phi(x)-\phi(0))\,\mathrm{d} x\right|\leq\varepsilon\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} \,dx
$$
Thus
$$
\Big|\frac{1}{\int\psi(nx)\,dx} \int\psi(nx)(\phi(x)-\phi(0))\,dx\Big|\leq \varepsilon
$$
for all $n$ large enough.
Notice that it is enough to consider $\phi\in\mathcal{C}_b(\mathbb{R})$. This means that
$\mu_n(dx)=\frac{1}{\int\psi(xn)\,dx}\psi(nx)\,dx$ converges weakly (as probability measures) to $\delta_0$.
