Question:
Let be $w(x)=1/x$ and $$w_\varepsilon(x)= \begin{cases} \frac{1}{x},&\quad |x| > \varepsilon\\ 0,&\quad |x| \leq \varepsilon\\ \end{cases}, $$ with $\varepsilon >0$.
Show that $\forall \phi \in \mathcal{D}(\mathbb{R}), \exists\; \lim_{\varepsilon \rightarrow 0^+} (w_\varepsilon,\phi) = \lim_{\varepsilon \rightarrow 0^+} \int_{\mathbb{R} \setminus [-\varepsilon,\varepsilon]} \frac{\phi(x)}{x}\; dx =: I_\varepsilon(x)$ and is this limit is finite.
Solution:
Let be $[-a,a]=\text{supp}(\phi)$ for an arbitrary $\phi$, and the notice that: $$\int_{\mathbb{R} \setminus [-\varepsilon,\varepsilon] \cap \text{supp}(\phi)} \frac{\phi(0)}{x}\;dx = \phi(0) \bigg( \int_{-a}^{-\varepsilon} + \int_a^\varepsilon \bigg) \frac{1}{x}\;dx = 0\; \quad (a)$$ because $\int_{-a}^{-\varepsilon} \frac{dx}{x} \overset{x=-y}{=} \int_\varepsilon^a \frac{dy}{y}$.
We also have that: $\lim_{x\rightarrow 0} \frac{\phi(x) - \phi(0)}{x} = \phi'(x) \quad (b)$
If from $I_\varepsilon(x)$ we substract $(a)$: $$ |g_\varepsilon(x)|=\bigg|\int_{\mathbb{R} \setminus [-\varepsilon,\varepsilon]} \frac{\phi(x)-\phi(0)}{x}\; dx \bigg|= \bigg|\int_{-a}^a \chi_\varepsilon(X) \frac{\phi(x) - \phi(0)}{x} \; dx\bigg|$$
And let $|g(x)| = \big|\frac{\phi(x)-\phi(0)}{x}\big| \in \mathcal{L}^1(-a,a)$, then we have that $|g_\varepsilon(x)| < |g(x)|$ because $|g_\varepsilon(x)| = |\chi_\varepsilon(x) g(x)|$.
Then $I_\varepsilon(x)$ is bounded, and from the dominated convergence theorem we can conclude that $\lim_{x\rightarrow 0^+} I_\varepsilon(x) = \int_{\mathbb{R}} \frac{\phi(x) -\phi(0)}{x}\; dx$.
I'm not sure about a couple of things:
1) It's is sufficient to show $(b)$ to ensure the function behave "well" around $0$ when integrated?
2) Is the proof complete or I'm missing some details?
EDIT:
Corrected $w_\epsilon(x)$ definition.
Corrected $w_\epsilon(x)$ definition, again, confused $<$ with $>$.
Let $\phi \in \mathcal D$ be given. Through a Maclaurin expansion of $\phi$ we can write $\phi(x) = \phi(0) + x \eta(x)$ where $\eta(x)$ is bounded as $x \to 0$.
Now take $\delta>0$.
Since $\eta$ is bounded at $0$ there is $a>0$ such that $\int_{|x|<a} |\eta(x)| \, dx < \delta$.
We clearly have $T_a := \int_{|x|>a} \frac{\phi(x)}{x} \, dx$ finite, so let's look at $\int_{\epsilon<|x|\leq a} \frac{\phi(x)}{x} \, dx$: $$ \int_{\epsilon<|x|\leq a} \frac{\phi(x)}{x} \, dx = \int_{\epsilon<|x|\leq a} \frac{\phi(0) + x \eta(x)}{x} \, dx = \phi(0) \int_{\epsilon<|x|\leq a} \frac{1}{x} \, dx + \int_{\epsilon<|x|\leq a} \eta(x) \, dx $$ Here, $\int_{\epsilon<|x|\leq a} \frac{1}{x} \, dx = 0$ since we integrate an odd function over a symmetric set.
Thus, $$ \left| \int_{\epsilon<|x|\leq a} \frac{\phi(x)}{x} \, dx \right| = \left| \int_{\epsilon<|x|\leq a} \eta(x) \, dx \right| \leq \int_{\epsilon<|x|\leq a} \left| \eta(x) \right| \, dx \leq \int_{|x|\leq a} \left| \eta(x) \right| \, dx < \delta $$ As this doesn't depend on $\epsilon$ we have $$\lim_{\epsilon \to 0} \left| \int_{\epsilon<|x|\leq a} \frac{\phi(x)}{x} \, dx \right| \leq \delta$$ so $$ \left| \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{\phi(x)}{x} \, dx - T_a \right| = \left| \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{\phi(x)}{x} \, dx - \int_{|x|>a} \frac{\phi(x)}{x} \, dx \right| \\ = \left| \lim_{\epsilon \to 0} \int_{\epsilon<|x|<a} \frac{\phi(x)}{x} \, dx \right| \leq \lim_{\epsilon \to 0} \left| \int_{\epsilon<|x|<a} \frac{\phi(x)}{x} \, dx \right| \leq \delta $$ i.e. $\lim_{\epsilon \to 0} \int_{|x|>\epsilon}$ exists and is finite.