= Solution
Introduce the <Cauchy approximate identity>
$$
\delta_\varepsilon(x)=\frac{\varepsilon}{\pi(\varepsilon^2+x^2)}.
$$
It is an <approximate identity> converging to the <Dirac delta function> $\delta$, and $g_\varepsilon=\delta_\varepsilon'$. For a smooth rapidly decaying <test function> $\phi$, <integration by parts> gives
$$
\int_{-\infty}^{\infty}\phi(x)g_\varepsilon(x)\,dx
=-\int_{-\infty}^{\infty}\phi'(x)\delta_\varepsilon(x)\,dx
\longrightarrow-\phi'(0).
$$
By the definition of the <derivative of the Dirac delta>, $\langle\delta',\phi\rangle=-\phi'(0)$. Therefore the limit as a <distribution> is
$$
\boxed{g_\varepsilon\longrightarrow\delta'}.
$$
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