$\lim_{\epsilon\to 0}\int_{\mathbb{R}\setminus(-\epsilon,\epsilon)}\frac 1 x \varphi(x) \, dx = \int_\mathbb{R}\frac 1 x \varphi_\text{odd}(x) \,dx $ Let $\varphi \in C^\infty_c(\mathbb{R})$. How can we prove that the following holds?
$$\lim_{\varepsilon\to 0}\int_\mathbb{R\smallsetminus(-\varepsilon,\varepsilon)} \frac 1 x \varphi(x) \, dx = \int_\mathbb{R} \frac 1 x \varphi_\text{odd}(x) dx, $$
where $$\varphi_\text{odd} = \frac{\varphi(x)-\varphi(-x)} 2$$

The support of $\varphi$ is within $[-R,R]$ for some $R>0$.
Then (change of variable + symmetry of $\varphi_\text{odd}$):
\begin{align}
\lim_\varepsilon \left(\int_{-R}^{-\varepsilon} + \int_{\varepsilon}^R\right) \frac{\varphi(x)} x \, dx & = \lim_\varepsilon \int_{\varepsilon}^R \frac{\varphi(x)-\varphi(-x)} x \, dx \\[12pt]
& = \int_0^\infty\frac{\varphi(x)-\varphi(-x)} x \, dx =\int_{\Bbb R} \frac{\varphi_\text{odd} (x)} x \, dx
\end{align}
