Help with $ \lim\limits_{t \to 0} \int_{-1}^1 \frac{t}{t^2+x^2} f(x)\ dx$ Show that if $f$ is continuous on $[-1,1]$, then
$$
\lim_{t \to 0} \int_{-1}^{1}\frac{t}{t^2+x^2}f(x)\,dx=\pi f(0)
$$
Any hints?
 A: Let us extend (by preserving continuity) the domain of $f$ by setting $f(x)=f(1)$ for any $x\geq 1$ and $f(x)=f(-1)$ for any $x\leq -1$. By the substitution $x=tz,\,dx=t\,dz$ we get:
$$ \int_{-1}^{1}\frac{t\,f(x)}{t^2+x^2}\,dx = \int_{-1/t}^{1/t}\frac{f(zt)}{1+z^2}\,dx =O(t)+\int_{-\infty}^{+\infty}\frac{f(zt)}{1+z^2}\,dz$$
and by the dominated convergence theorem:
$$ \lim_{t\to 0^+}\int_{-1}^{1}\frac{t\,f(x)}{t^2+x^2}\,dx = \lim_{t\to 0^+}\int_{-\infty}^{+\infty}\frac{f(zt)}{1+z^2}\,dz = \int_{-\infty}^{+\infty}\frac{f(0)}{1+z^2}\,dz = \pi\,f(0) $$
as wanted. The $O(t)$ term comes from
$$ \int_{1/t}^{+\infty}\frac{1}{1+z^2}\,dz = \arctan(t) = O(t).$$
A: $\boldsymbol{\delta}$-$\boldsymbol{\epsilon}$ Approach
Choose an $\epsilon\gt0$. Pick a $0\lt\delta\lt1$ so that $|x|\le\delta\implies|f(x)-f(0)|\le\epsilon$. Pick an $M$ so that $\int_{|x|\ge M}\frac1{1+x^2}\,\mathrm{d}x\le\frac{\epsilon}{\max\limits_{|x|\le1}|f(x)|}$. Then for $t\lt \delta/M$,
$$
\begin{align}
&\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|\\
&=\left|\,\int_{-1/t}^{1/t}\frac1{1+x^2}f(xt)\,\mathrm{d}x-\int_{-\infty}^\infty\frac1{1+x^2}f(0)\,\mathrm{d}x\,\right|\\
&\le\int_{|x|\ge1/t}\frac1{1+x^2}|f(0)|\,\mathrm{d}x+\int_{|x|\lt1/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x\\
&\le\color{#C00}{\int_{|x|\ge1/t}\frac1{1+x^2}|f(0)|\,\mathrm{d}x}+\color{#090}{\int_{\delta/t\le|x|\lt1/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x}\\
&+\color{#00F}{\int_{|x|\lt\delta/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x}\\[6pt]
&\le\color{#C00}{\epsilon}+\color{#090}{2\epsilon}+\color{#00F}{\pi\epsilon}
\end{align}
$$
Thus, for any $\epsilon\gt0$
$$
\lim_{t\to0}\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|\le(3+\pi)\epsilon
$$
which means that
$$
\lim_{t\to0}\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|=0
$$

Dominated Convergence Approach
$[|xt|\le1]$, where $[\cdots]$ are Iverson brackets, is the characteristic function of $\left[-\frac1t,\frac1t\right]$. Then
$\left|\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}\right|\le\overbrace{\max\limits_{|u|\le1}|f(u)|\frac1{1+x^2}}^{L^1}$ and for all $x\in\mathbb{R}$, $\lim\limits_{t\to0}\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}=\color{#090}{\frac{f(0)}{1+x^2}}$. Therefore,
$$
\begin{align}
\lim_{t\to0}\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x
&=\lim_{t\to0}\int_{-1/t}^{1/t}\frac1{1+x^2}f(xt)\,\mathrm{d}x\\
&=\lim_{t\to0}\int_{-\infty}^\infty\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}\,\mathrm{d}x\\[3pt]
&=\int_{-\infty}^\infty\color{#090}{\frac{f(0)}{1+x^2}}\,\mathrm{d}x\\[9pt]
&=\pi f(0)
\end{align}
$$
