Show that $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}y_\epsilon(x)=\delta(x)$ For $y_\epsilon(x)=\frac{1}{1-e^{-\frac{2}{\epsilon}}}(e^{-\frac{x}{\epsilon}}-e^{-\frac{2-x}{\epsilon}}),\ x\in[0,1]$ prove it converges to Dirac's delta as ${\epsilon\rightarrow0}$.
Answer:
It is sufficient to prove $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_0^x y_\epsilon(u)\phi(u)\,du=\phi(0)$, for any $\phi\in C_c(\mathbb{R})$.
The $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_0^x y_\epsilon(u)\, du=1$, after some calculations. Therefore I can take $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_0^x y_\epsilon[\phi(x)-\phi(0)]\,dx$.
But can I use continuity? How can I know that $\lvert x-0\rvert<\delta$?
Second Thought:
Since $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_0^x y_\epsilon(u)\, du=1$, change variables as $z=\frac{u}{\epsilon}$ and $\lim_{\epsilon\rightarrow0}\int_0^{x/\epsilon}y_\epsilon(z\epsilon)\ dz=\int_0^{x/\epsilon}y_\epsilon(0)\ dz=1$.
Then use Dominated Convergence:
$\lim_{\epsilon\rightarrow0}\int_0^{x/\epsilon}y_\epsilon(z\epsilon)\phi(ze)\ dz=\int_0^{x/\epsilon}y_\epsilon(0)\phi(0)\ dz=\phi(0)$
 A: You can use the  change of variables you suggest along the mean value theorem and some integration by parts:
$$
y_\varepsilon(x)=\frac{1}{1-e^{-\frac{2}{\epsilon}}}\Big(e^{-x\varepsilon^{-1}}-e^{-2\varepsilon^{-1}}e^{x\varepsilon^{-1}}\Big)\mathbb{1}_{[0,1]}(x)$$
For any $\phi\in\mathcal{C}^\infty_c(\mathbb{R})$
\begin{align}
\frac{1}{\varepsilon}\int_\mathbb{R} y_{\varepsilon}(x)\phi(x)&=\frac{1}{1-e^{-\frac{2}{\epsilon}}}\int^{\varepsilon^{-1}}_{0}(e^{-u}-e^{-2\varepsilon^{-1}}e^u\big)\phi(\varepsilon u)\,du\\
&=\frac{1}{1-e^{-\frac{2}{\epsilon}}}\Big(\int^{\varepsilon^{-1}}_{0}(e^{-u}-e^{-2\varepsilon^{-1}}e^u\big)\big(\phi(\varepsilon u)-\phi(0)\Big)\,du \\
&\quad +\phi(0)\int^{\varepsilon^{-1}}_{0}e^{-u}-e^{-2\varepsilon^{-1}}e^u\,du\Big)
\end{align}
The term
$$
\frac{1}{1-e^{-\frac{2}{\epsilon}}}\phi(0)\int^{\varepsilon^{-1}}_{0}e^{-u}-e^{-2\varepsilon^{-1}}e^u\,du=\phi(0)\frac{\big(1-e^{-\varepsilon^{--1}}\big)^2}{1-e^{-\frac{2}{\epsilon}}}\xrightarrow{\varepsilon\rightarrow0+}\phi(0)
$$
The term
\begin{align}
\frac{\int^{\varepsilon^{-1}}_{0}(e^{-u}-e^{-2\varepsilon^{-1}}e^u\big)\big(\phi(\varepsilon u)-\phi(0)\Big)\,du}{1-e^{-2\varepsilon^{-1}}}&\leq \frac{\|\phi'\|_\infty\varepsilon}{1-e^{-\frac{2}{\epsilon}}}\int^{\varepsilon^{-1}}_0u\big(e^{-u}+e^{-2\varepsilon^{-1}}e^u\big)\,du\\
&=\frac{\|\phi'\|_\infty\varepsilon}{1-e^{-\frac{2}{\epsilon}}}\Big(1-\varepsilon^{-1}e^{-\varepsilon^{-1}}-e^{-\varepsilon^{-1}}\\
&\quad\quad+ \varepsilon^{-1}e^{-\varepsilon^{-1}}-e^{-\varepsilon^{-1}}+e^{-2\varepsilon^{-1}}\Big)
\Big)
\xrightarrow{\varepsilon\rightarrow0+}0
\end{align}
