I am asked to prove that $f_t = \frac{x}{x^2+t}$ convergences in the sense of distributions to the Principal Value distribution when $t>0$ and $t \to 0$.
The principal value distribution $T$ is given by $T(h)=\int_0^\infty \frac{h(x)-h(-x)}{x}d x$.
Let $h \in \mathcal{D} (\mathbb{R})$.
The problem is that when I do the calculation of $(T-T_{f_t})(h)$ I end up with $$t \int_0^\infty \frac{h(x)-h(-x)}{x^2+t}dx=\int_0^\infty \frac{2h'(0)}{(x/\sqrt{t})^2+1}dx + \int_0^\infty t \frac{o(x^2)}{x^2+t}dx.$$
It is not an issue to show that the second integral converges to $0$ since $o(x^2)/x$ is bounded in $[0,1]$ (which we need after applying $x^2 + t\ge2xt$).
On the other hand, on $]1, \infty[$, $o(x^2)$ is bounded so we can apply the dominated convergence theorem.
As of the first one, one would get $c \cdot h'(0)\cdot \sqrt t.$ Is this enough to deduce the convergence?
Answering this question so that it does not appear as unanswered:
As @TZakrevskiy noted, the solution is correct.