Let $c > 0$ be such that $a < -c < c < b$. Observe that $\lim_{\epsilon \rightarrow 0} \bigg(\int_a^{-c}{f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx + \int_{c}^b {f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx\bigg) = \int_a^{-c}{f(x) \over x} \,dx + \int_{c}^b {f(x) \over x}\,dx $ (We can just plug in $\epsilon = 0$ when we take limits here since the denominators are never zero.) So it suffices to replace $a$ by $-c$ and $b$ by $c$; in other words it suffices to show that $\lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx = p.v. \int_{-c}^c {f(x) \over x}$ Since ${\displaystyle {f(0) \over x}{x^2 \over x^2 + \epsilon^2}}$ is an odd function and the domain of integration is symmetric about $x = 0$, you can subtract it from the integrand on the left-hand side without changing the limit. So this limit is equal to $\lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) - f(0) \over x} {x^2 \over x^2 + \epsilon^2}\,dx$ Note that |{f(x) - f(0) \over x}| < \sup_x|f'(x)|, and that $|{x^2 \over x^2 + \epsilon^2}| < 1$. So the integrand above is uniformly bounded and one can just take the limit as $\epsilon$ goes to zero, using the dominated convergence theorem for example. The result is $\int_{|x| < c} {f(x) - f(0) \over x} \,dx$ $= \lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) - f(0) \over x} \,dx$ We can now go in the reverse direction, and add the odd function ${f(0) \over x}$ to the integrand, and the above is equal to $= \lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) \over x} \,dx$ $= p.v. \int_{-c}^c {f(x) \over x}\,dx$