Show $\lim_{n\to \infty} \int_{-1}^1 \frac n 2 e^{-n|x|}f(x)dx=f(0)$ if $f: [-1,1] \to \Bbb{R}$ is continuous. For a continuous function $f:[-1,1] \to \Bbb R$, show that $$\lim_{n\to \infty} \int_{-1}^1 \frac n 2 e^{-n|x|}f(x)dx=f(0)$$
I showed this by given $\epsilon>0$, choose $\delta>0$ such that $|x|<\delta \implies |f(x)-f(0)|<\epsilon$. Then divide the integral into $[-\delta,\delta]$ and $[-1,1]-[-\delta,\delta]$ (as in this answer), and I think it works, although I will not write the details in here. But I'm curious that if there is a better proof. My proof seems elementary and a little messy.
 A: Using the substitution $y = nx$, we arrive at
$$
\frac{1}{2}\int_{-n}^n e^{-|y|} f(y/n)\,dy = \frac{1}{2}\int \chi_{[-n,n]}(y) e^{-|y|} f(y/n)\, dy,
$$
where $\chi_{[-n,n]}$ is the characteristic function/indicator function of the interval $[-n,n]$ and we have extended $f$ by $0$ to all of $\Bbb{R}$.
Now the integrand converges pointwise to $e^{-|y|}\cdot f(0)$, and is dominated (in absolute value) by the integrable function $y\mapsto e^{-|y|} \cdot \max_{x\in [-1,1]}|f(x)|$. Use the dominated convergence theorem (I hope you know it, otherwise this answer will not help you much) to conclude the proof.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\lim_{n\ \to\ \infty}\ \int_{-1}^{1}
     {n \over 2}\,\expo{-n\verts{x}}\fermi\pars{x}\,\dd x = \fermi\pars{0}:\
     {\large ?}.\qquad\fermi:\bracks{-1,1}\ \to\ {\mathbb R}\,,\quad\fermi}$
is continuous.

\begin{align}
\mbox{Note that}&\quad\dsc{\int_{-1}^{1}{n \over 2}\,\expo{-n\verts{x}}\fermi\pars{x}\,\dd x}
=\half\int_{-1}^{1}n\expo{-n\verts{x}}
\bracks{\fermi\pars{x} - \fermi\pars{0}}\,\dd x + \fermi\pars{0}
\\[5mm]&=\half\int_{-n}^{n}\expo{-\verts{x}}
\bracks{\fermi\pars{x \over n} - \fermi\pars{0}}\,\dd x + \fermi\pars{0}
\end{align}


The whole problem is reduced to prove that
  $\ds{\lim_{n\ \to\ \infty}\ \half\int_{-n}^{n}\expo{-\verts{x}}
\bracks{\fermi\pars{x \over n} - \fermi\pars{0}}\,\dd x = \dsc{0}}$.


Given $\ds{\epsilon > 0, \exists\ \delta > 0\ \mid\
\braces{~0 < \verts{x \over n} < \delta\ \imp\
\verts{\fermi\pars{x \over n} - \fermi\pars{0}} < 2\epsilon~}}$. Then,
$\ds{\exists\ N \equiv \floor{\verts{x} \over \delta}\ \mid\
\bracks{~n > N\ \imp\ \verts{\fermi\pars{x \over n} - \fermi\pars{0}} < 2\epsilon~}}$. It's true that

$$
\verts{\half\int_{-n}^{n}\expo{-\verts{x}}
\bracks{\fermi\pars{x \over n} - \fermi\pars{0}}\,\dd x}
<\pars{\half\int_{-n}^{n}\expo{-\verts{x}}\,\dd x}\epsilon < \epsilon\ \quad\mbox{when}\quad n > N
$$


This completes the proof.

