Solving $\lim_{t\rightarrow 0}\frac{1}{\pi}\int_{-1}^1 f(x) \frac{t}{t^2+x^2} dx =f(0)$ without dominated convergence I have to prove the following:
$$
\lim_{t\rightarrow 0}\frac{1}{\pi}\int_{-1}^1 f(x) \frac{t}{t^2+x^2} dx =f(0)
$$
Assuming continuity of $f(x)$ for every real number. I tried change of variable with $x=t \tan(\theta)$ and $u=\frac{x}{t}$ but none of them take me to some known place. 
Actually, when I try the second change of variable it looks like Dirac delta function. But maybe I did something wrong or going in the wrong way.
I know that it can be solved with Dominated Convergence Theorem, but I'm not allowed to use it.
 A: There is a commmon technique for these sorts of approximation to the identity problems: Break the integral up into two pieces, one of which gets small and the other, using the continuity of $f$, you can show converges to $f(0)$.
Fix $\epsilon>0$. Find $\delta>0$ so that for $|x|<\delta$, 
$$
f(0)-\epsilon<f(x)<f(0)+\epsilon
$$
Then 
$$
(f(0)-\epsilon)\int_{-\delta}^\delta\frac{t}{t^2+x^2}\mathrm dx <\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}<(f(0)+\epsilon)\int_{-\delta}^\delta\frac{t}{t^2+x^2}\mathrm dx 
$$
Now you can do the substitution you mention on the integrals on left and right of the inequalities. Namely,
$$
\int_{-\delta}^\delta\frac{t}{t^2+x^2}\mathrm dx =\int_{\arctan(-\delta/t)}^{\arctan(\delta/t)}\mathrm d\theta=2\arctan(\delta/t)\to \pi
$$
as $t\to 0^+$, which is definitely necessary (the limit does not exist from both sides). So, 
$$
\lim_{t\to 0^+}(f(0)-\epsilon)\int_{-\delta}^\delta\frac{t}{t^2+x^2}\mathrm dx <\lim_{t\to 0^+}\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}\mathrm dx<\lim_{t\to 0^+}(f(0)+\epsilon)\int_{-\delta}^\delta\frac{t}{t^2+x^2}\mathrm dx\\
\implies\pi (f(0)-\epsilon) <\lim_{t\to 0^+}\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}\mathrm dx<\pi(f(0)+\epsilon)\\
$$
so that $f(0)-\epsilon<\lim_{t\to 0^+}\frac{1}{\pi}\lim_{t\to 0^+}\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}\mathrm dx<f(0)+\epsilon$. Since $\epsilon>0$ was arbitrary, $\lim_{t\to 0^+}\frac{1}{\pi}\lim_{t\to 0^+}\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}\mathrm dx=f(0)$.
Now, for the other, "small," piece,
$$
\left|\int_{[-1,1]\setminus (-\delta,\delta)}f(x)\frac{t}{t^2+x^2}\mathrm dx\right|\leq \sup_{y\in[-1,1]}|f(y)|\int_{[-1,1]\setminus (-\delta,\delta)}\frac{|t|}{t^2+x^2}\mathrm dx\\
\leq 
\sup_{y\in[-1,1]}|f(y)|\frac{|t|}{t^2+\delta^2}\int_{[-1,1]\setminus (-\delta,\delta)}\mathrm dx \\
=\sup_{y\in[-1,1]}|f(y)|\frac{|t|}{t^2+\delta^2}2(1-\delta)
$$
which goes to $0$ with $t\to 0$.
So, putting it all together, 
$$
\lim_{t\to 0^+}\frac{1}{\pi}\int_{-1}^1f(x)\frac{t}{t^2+x^2}\mathrm dx\\
=\lim_{t\to 0^+}\frac{1}{\pi}\int_{-\delta}^\delta f(x)\frac{t}{t^2+x^2}\mathrm dx+\lim_{t\to 0^+}\frac{1}{\pi}\int_{[-1,1]\setminus (-\delta,\delta)} f(x)\frac{t}{t^2+x^2}\mathrm dx\\
=f(0)
$$
A: Since $f$ is continuous, then for all $\varepsilon>0$ there exists a $\delta>0$ such that whenever $|x|<\delta$, $|f(x)-f(0)|<\varepsilon$.
With such a chosen $\varepsilon>0$ and associated $\delta>0$, we fix $\delta$ and write
$$\begin{align}
\int_{-1}^1\,f(x)\,\frac{t}{t^2+x^2}\,dx&=\int_{-1}^1\,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx+f(0)\int_{-1}^1 \frac{t}{t^2+x^2}\,dx\\\\
&=\int_{-1}^1\,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx+2\arctan(1/t)f(0)\\\\
&=2\arctan(1/t)f(0)+\int_{|x|<\delta} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx\\\\
&+\int_{\delta<|x|<1} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx\tag1
\end{align}$$
As $t\to 0^+$, the first term on the right-hand side of $(1)$ approaches $\pi f(0)$ while the third term approaches $0$.  We next have a look at the second term on the right-hand side of $(1)$.  Proceeding we have the estimates
$$\begin{align}
\left|\int_{|x|<\delta} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx\right|&\le 2\varepsilon \int_0^\delta \frac{|t|}{t^2+x^2}\,dx\\\\
&=2\varepsilon \arctan(\delta/|t|)\\\\
&\le \pi \varepsilon
\end{align}$$
Putting it all together we find that 
$$\lim_{t\to0^+}\frac1\pi\int_{-1}^1\,f(x)\,\frac{t}{t^2+x^2}\,dx=f(0)$$
as was to be shown!
A: I assume that $t>0$. Note that
$$
\frac{1}{\pi }\int_{ - 1}^1 {f(x)\frac{t}{{t^2  + x^2 }}dx}  = \frac{1}{\pi }\int_{ - 1/t}^{1/t} {\frac{{f(tu)}}{{1 + u^2 }}du}  = \frac{1}{\pi }\int_{ - \infty }^{ + \infty } {\frac{{f(tu)\chi _{\left( { - 1/t,1/t} \right)} (u)}}{{1 + u^2 }}du} ,
$$
where $\chi$ is the characteristic function of the given interval. Since
$$
\left| {f(tu)\chi _{\left( { - 1/t,1/t} \right)} (u)} \right| \le \mathop {\max }\limits_{\left| s \right| \le 1} \left| {f(s)} \right| <  + \infty ,
$$
we can use the dominated convergence theorem and obtain
\begin{align*}
& \mathop {\lim }\limits_{t \to 0+} \frac{1}{\pi }\int_{ - 1}^1 {f(x)\frac{t}{{t^2  + x^2 }}dx}  = \mathop {\lim }\limits_{t \to 0+} \frac{1}{\pi }\int_{ - \infty }^{ + \infty } {\frac{{f(tu)\chi _{\left( { - 1/t,1/t} \right)} (u)}}{{1 + u^2 }}du} 
\\ & = \frac{1}{\pi }\int_{ - \infty }^{ + \infty } {\frac{{\mathop {\lim }\limits_{t \to 0+} f(tu)\chi _{\left( { - 1/t,1/t} \right)} (u)}}{{1 + u^2 }}du}  = \frac{1}{\pi }\int_{ - \infty }^{ + \infty } {\frac{{f(0)}}{{1 + u^2 }}du} \\ & = f(0)\frac{1}{\pi }\int_{ - \infty }^{ + \infty } {\frac{{du}}{{1 + u^2 }}}  = f(0).
\end{align*}
