$Q_t=\frac{1}{\pi} \frac{t}{t^2 + x^2}.$ Then $\lim_{t\to 0} Q_t = \frac{1}{\pi} \text{p.v.} (\frac{1}{x})$ This is a theorem from a class note.
Let $Q_t=\frac{1}{\pi} \frac{t}{t^2 + x^2}.$ Then $\lim_{t\to 0} Q_t = \frac{1}{\pi} \text{p.v.} (\frac{1}{x})$ in the sense of a distribution. In other words, 
$$\lim_{t\to 0}[\frac{1}{\pi} \int \frac{t}{t^2+x^2}\phi(x)dx-\frac{1}{\pi} \int_{|x|>t}\frac{\phi(x)}{x}dx]=0.$$where $\phi$ is a Schwartz function or a $C_c^\infty$ function. 
Proof. We can split it into two parts. First, $$\lim_{t\to 0}\int_{|x|<t}\frac{t}{t^2+x^2}\phi(x) dx = \lim_{t\to 0}\int_{|z|<1} \frac{t}{t^2+t^2z^2}\phi(tz)tdz = \lim_{t\to 0}\int_{|z|<1}\frac{1}{1+z^2}\phi(tz)dz$$ 
and 
$$\lim_{t\to 0}\int_{|x|>t} [\frac{t}{t^2 + x^2}-\frac{1}{x}]\phi(x)dx=\lim_{t\to 0}\int_{|z|>1} [\frac{t}{t^2(1+z^2)}-\frac{1}{tz}]\phi(tz)tdz = \lim_{t\to 0}\int_{|z|>1}[\frac{1}{1+z^2}-\frac{1}{z}]\phi(tz)dz$$
Now the note says that putting these two back together and using the dominated convergence theorem, we get $0$. But how is this so? Using DCT gives us $\phi(0)$ on both integrals and ignoring it we get $\int \frac{1}{1+z^2}+\int_{|z|>1}\frac{1}{z} dz$ but the second integral is not even defined so how do we get $0$? I would greatly appreciate any help.
 A: I know that this is a rather old question but it is still unanswered.
Is the claim really true? I instead think that $\lim_{t \to 0} Q_t = \pi \delta$.
Here's my reasoning:
I cannot see a need to split the integral. The substitution $x = tz$ makes
$$\langle Q_t, \phi \rangle = \int Q_t(x) \, \phi(x) \, dx = \int \frac{t}{t^2+x^2} \, \phi(x) \, dx
= \int \frac{1}{1+z^2} \, \phi(tz) \, dz.$$
Now, $\frac{1}{1+z^2} \in L^1(\mathbb R)$ and the factor $\phi(tz)$ doesn't make this worse. In fact we have $\left| \frac{1}{1+z^2} \, \phi(tz) \right| \leq \frac{1}{1+z^2} \|\phi\|_\infty \in L^1(\mathbb R).$ We therefore can use the dominated convergence theorem to conclude that
$$\lim_{t \to 0} \, \langle Q_t, \phi \rangle = \lim_{t \to 0} \int \frac{1}{1+z^2} \, \phi(tz) \, dz = \int \frac{1}{1+z^2} \, \lim_{t \to 0} \phi(tz) \, dz \\ = \int \frac{1}{1+z^2} \, \phi(0) \, dz = \int \frac{1}{1+z^2} \, dz \, \phi(0) = \pi \, \phi(0) = \langle \pi \delta, \phi \rangle$$
Thus, $\lim_{t \to 0} Q_t = \pi \delta.$
