Question: What is the limit of $2x/(x^2+\epsilon^2)$ in the distributional sense, as $\epsilon\to 0$?

For any $\phi\in C\_c^\infty(\Bbb R)$, $\int \frac{2x}{x^2+\epsilon^2}\phi(x)dx\to what$? as $\epsilon\to 0$? then we have the limit. 

If the limit do not have an explicit formula, how can we show the convergence of $2x/(x^2+\epsilon^2)$ in the distributional sense.

Answer: 
A rigorous treatment isn't very hard:

Let $u\_\epsilon(x) = \frac{2x}{x^2+\epsilon^2}$ and take $\phi \in C\_c^\infty(\mathbb R)$. Then
$$
\langle u\_\epsilon, \phi \rangle
= \int \frac{2x}{x^2+\epsilon^2} \, \phi(x) \, dx
= \int \left( \ln |x^2+\epsilon^2| \right)' \, \phi(x) \, dx \\
= \{ \text{ partial integration } \} 
= - \int \ln |x^2+\epsilon^2| \, \phi'(x) \, dx \\
= \{ \text{ $\ln |x^2+\epsilon^2|$ converges to integrable function $\ln |x^2|$ } \} \\
\to - \int \ln |x^2| \, \phi'(x) \, dx 
= - 2 \int \ln |x| \, \phi'(x) \, dx 
= - 2 \, \langle \ln |x|, \phi' \rangle \\
= \{ \text{ definition of distributional derivative } \} \\
= 2 \, \langle \left( \ln |x| \right)', \phi \rangle 
= 2 \, \langle \operatorname{pv} \frac{1}{x}, \phi \rangle 
= \langle 2 \operatorname{pv} \frac{1}{x}, \phi \rangle
$$
WARNING: The following contains the kind of „physicist’s“ reasoning that some viewers might find offensive

$$\lim \_{\epsilon \to 0} \int \frac{2x}{x^2+\epsilon^2} \phi(x) \mathrm{d}x = 
 \lim \_{\epsilon \to 0} 2 \int \frac{x^2}{x^2+\epsilon^2} \frac{\phi(x)}{x} \mathrm{d}x = 
2 \mathcal{P} \int \frac{\phi(x)}{x} $$

The principal value integral arises, in intuitive terms, as the term $\frac{x^2}{x^2+\epsilon^2} $ tends to $0$ for $x<< \epsilon$, and to $1$ for $x >> \epsilon$