$f \in C[-1,1]$, Prove ${\lim_{h \to 0^+}}{\int_{-1}^1 \frac{h}{h^2+x^2}f(x)\,dx} = \pi f(0)$ I took a look at the special situation that $f=1$,
$${\lim_{h \to 0^+}}{\int_{-1}^1 \frac{h}{h^2+x^2}f(x)\,dx} ={\lim_{h \to 0^+}}{\int_{-1}^1 \frac{h}{h^2+x^2}\,dx} = \left.{\lim_{h \to 0^+}} \arctan{\frac{x}{h}} \right|_{-1}^1 = \pi $$
but I don't know how to find the next step.
 A: Let $I_h$ denote your intgeral. By setting $x=th$ we get
$$
I_h=\int^{1/h}_{-1/h}\frac{1}{1+t^2}f(th)\,dt
$$
Using Dominated convergence the assertion is trivial, but let us avoid such heavy tool. We know that $f$ is continuous on $[-1,1]$ a compact interval, so $f$ is uniformly continuous. That means there exists a non-decreasing continuous function $\omega:[0,\infty)\to[0,\infty)$ such that $\omega(0)=0$ and 
\begin{align*}
|f(x)-f(y)|\leq \omega(|x-y|)
\end{align*}
We will use this later, but first write
\begin{align}|I_h-\pi f(0)|=\left|\int^{1/h}_{-1/h}\frac{1}{1+t^2}f(th)\,dt-\int^\infty_{-\infty} \frac{f(0)}{1+t^2}\,dt \right|\end{align}
Now we will split this into different pieces
\begin{align*}
|I_h-\pi f(0)|=\left|\color{blue}{\int^{1/\sqrt h}_{-1/\sqrt h}}+\color{red}{\int^{1/ h}_{1/\sqrt h}}+\color{purple}{\int^{-1/\sqrt h}_{-1/h}}\frac{1}{1+t^2}f(th)\,dt-\left(\color{blue}{\int^{1/\sqrt h}_{-1/\sqrt h}}+\color{red}{\int^\infty_{1/\sqrt h}} +\color{purple}{\int^{-1/\sqrt h}_{-\infty}}\frac{f(0)}{1+t^2}\,dt \right)\right|
\end{align*}
The reason we have colored the integrals is because we can take the ones with the same colors together. We take them together through the triangle inequality as follows:
\begin{align}
|I_h-\pi f(0)|\leq \color{blue}{\int^{1/\sqrt h}_{-1/\sqrt h}\frac{|f(th)-f(0)|}{1+t^2}\,dt}+\color{red}{ \int^\infty_{1/\sqrt h} \frac{|f(th)\chi_{\{th\leq 1\}}-f(0)|}{1+t^2}\,dt} + \color{purple}{\int^{-1/\sqrt h}_{-\infty} \frac{|f(th)\chi_{\{-1\leq th\}}-f(0)|}{1+t^2}\,dt} 
\end{align}
where we have set $\chi$ to make sure the term is well defined. First we handle the blue ones. Recall that the function $\omega$ is non-decreasing so
$$\color{blue}{\int^{1/\sqrt h}_{-1/\sqrt h} \frac{|f(th)-f(0)|}{1+t^2}\,dt\leq \int^{1/\sqrt h}_{-1/\sqrt h} \frac{\omega(|th|)}{1+t^2}\,dt \stackrel{ |th|\leq \sqrt h }{\leq}  \int^{1/\sqrt h}_{-1/\sqrt h}\frac{\omega(\sqrt h)}{1+t^2}\,dt\leq \pi\omega(\sqrt h) }$$
The red and purple ones can be handled with the same way. Notice that due to continuity of $f$ and compactness of $[-1,1]$ we have the existence of $M>0$ such that $|f(x)|\leq M$ for all $x\in[-1,1]$. Therefore we get
\begin{align}
\color{red}{ \int^\infty_{1/\sqrt h} \frac{|f(th)\chi_{\{th\leq 1\}}-f(0)|}{1+t^2}\,dt\leq 2M \int^\infty_{1/\sqrt h} \frac{1}{1+t^2}\,dt = 2M \left( \frac \pi 2 -\arctan(1/\sqrt h)\right) }
\end{align}
and
\begin{align}
\color{purple}{\int^{-1/\sqrt h}_{-\infty} \frac{|f(th)\chi_{\{-1\leq th\}}-f(0)|}{1+t^2}\,dt \leq 2M \left(\arctan(-1/\sqrt h)+\frac \pi 2 \right)}
\end{align}
Therefore
\begin{align}
|I_h-\pi f(0)|\leq \pi\omega(\sqrt h) + 2M (\pi -2 \arctan(1/\sqrt h)) \rightarrow 0 \ \ \text{ as } h\to 0^+
\end{align}
which proves the claim.
BTW you can avoid to use the function $\omega$. It is useful, but it might be the case that you did not know about. You can use uniform continuity "manually" in that case...
A: Note that
$$\begin{aligned}
&\quad\int_{-1}^1\frac{h}{h^2+x^2}f(x)\,dx-2\arctan\frac{1}{h}f(0)\\&=\int_{-1}^1\frac{h}{h^2+x^2}(f(x)-f(0))\,dx\\&=\int_{|x|<\delta}\frac{h}{h^2+x^2}(f(x)-f(0))\,dx+\int_{1\geq|x|\geq\delta}\frac{h}{h^2+x^2}(f(x)-f(0))\,dx,
\end{aligned}$$
where $\delta>0$ is such that $|f(x)-f(0)|<\varepsilon$ whence $|x|<\delta$ (Since $f(x)$ is continous). Then, we have
$$\begin{aligned}
&\quad\left|\int_{-1}^1\frac{h}{h^2+x^2}f(x)\,dx-2\arctan\frac{1}{h}f(0)\right|\\&\leq
\int_{|x|<\delta}\frac{h}{h^2+x^2}|f(x)-f(0)|\,dx+\int_{1\geq|x|\geq\delta}\frac{h}{h^2+x^2}|f(x)-f(0))|\,dx\\&\leq\varepsilon\int_{-1}^1\frac{h}{h^2+x^2}\,dx+2\sup|f|\int_{1\geq|x|\geq\delta}\frac{h}{h^2+x^2}\,dx\\&=2\varepsilon\arctan\frac{1}{h}+4\sup|f|\left(\arctan\frac{1}{h}-\arctan{\frac{\delta}{h}}\right)\\&\leq\pi\varepsilon,\,\,\, as\quad h\to 0. 
\end{aligned}$$
Since $\varepsilon$ is arbitrary, we know
$$\lim_{h\to 0}\int_{-1}^1\frac{h}{h^2+x^2}f(x)\,dx=\lim_{h\to 0}2\arctan\frac{1}{h}f(0)=\pi f(0).$$
A: Let $\epsilon\gt0$. Since $f$ is continuous, there is some $\delta\gt0$ so that $|x-0|\le\delta\implies|f(x)-f(0)|\le\frac\epsilon{3\pi}$.
Furthermore, since $f$ is continuous on $[-1,1]$, it is bounded, say $|f(x)|\le M$.
Note that
$$
\begin{align}
\int_{-\delta}^\delta\frac{h}{h^2+x^2}\,\mathrm{d}x
&=2\tan^{-1}\left(\frac\delta{h}\right)\tag{1a}\\
&=\pi-2\tan^{-1}\left(\frac{h}\delta\right)\tag{1b}
\end{align}
$$
Explanation:
$\text{(1a)}$: integration
$\text{(1b)}$: $\tan(\pi/2-x)=1/\tan(x)$
Since $\int_{-\infty}^\infty\frac{h}{h^2+x^2}\,\mathrm{d}x=\pi$, $(1)$ says that
$$
\int_{|x|\gt\delta}\frac{h}{h^2+x^2}\,\mathrm{d}x=2\tan^{-1}\left(\frac{h}\delta\right)\tag2
$$
Therefore, with $h\le\delta\tan\left(\frac\epsilon{12M}\right)$
$$
\begin{align}
\left|\,\int_{-1}^1(f(x)-f(0))\,\frac{h}{h^2+x^2}\,\mathrm{d}x\,\right|
&\le\left|\,\int_{-\delta}^\delta(f(x)-f(0))\,\frac{h}{h^2+x^2}\,\mathrm{d}x\,\right|\\
&+\left|\,\int_{\delta\lt|x|\le1}(f(x)-f(0))\,\frac{h}{h^2+x^2}\,\mathrm{d}x\,\right|\tag{3a}\\[3pt]
&\le\frac\epsilon3+\frac\epsilon3\tag{3b}
\end{align}
$$
Explanation:
$\text{(3a)}$: triangle inequality
$\text{(3b)}$: on $[-\delta,\delta]$, $|f(x)-f(0)|\le\frac\epsilon{3\pi}$ and $\int_{-\infty}^\infty\frac{h}{h^2+x^2}\,\mathrm{d}x=\pi$
$\phantom{\text{(3b):}}$ for $\delta\lt|x|\le1$, $|f(x)-f(0)|\le2M$ and $\int_{|x|\gt\delta}\frac{h}{h^2+x^2}\,\mathrm{d}x=2\tan^{-1}\left(\frac{h}\delta\right)\le\frac\epsilon{6M}$
If we also ensure that $\delta\le\tan\left(\frac\epsilon{6M}\right)$,
$$
\begin{align}
\left|\,\int_{-1}^1f(0)\,\frac{h}{h^2+x^2}\,\mathrm{d}x-\pi f(0)\,\right|
&\le2M\tan^{-1}(\delta)\tag{4a}\\[3pt]
&\le\frac\epsilon3\tag{4b}
\end{align}
$$
Explanation:
$\text{(4a)}$: $|f(0)|\le M$ and $\text{(1b)}$ with $\delta=1$
$\text{(4b)}$: $\delta\le\tan\left(\frac\epsilon{6M}\right)$
Thus, with $0\lt\delta\le\tan\left(\frac\epsilon{6M}\right)$ so that $|x-y|\le\delta\implies|f(x)-f(y)|\le\frac\epsilon{3\pi}$ and $h\le\delta\tan\left(\frac\epsilon{12M}\right)$, we have
$$
\begin{align}
\left|\,\int_{-1}^1f(x)\,\frac{h}{h^2+x^2}\,\mathrm{d}x-\pi f(0)\,\right|
&\le\left|\,\int_{-1}^1(f(x)-f(0))\,\frac{h}{h^2+x^2}\,\mathrm{d}x\,\right|\\
&+\left|\,\int_{-1}^1f(0)\,\frac{h}{h^2+x^2}\,\mathrm{d}x-\pi f(0)\,\right|\tag{5a}\\[6pt]
&\le\epsilon\tag{5b}
\end{align}
$$
Explanation:
$\text{(5a)}$: triangle inequality
$\text{(5b)}$: apply $(3)$ and $(4)$
Finally, $(5)$ says that
$$
\lim_{h\to0^+}\int_{-1}^1f(x)\,\frac{h}{h^2+x^2}\,\mathrm{d}x=\pi f(0)\tag6
$$
