I want to find second derivative of $P.V. \frac{1}{x}$. And solution is $$\langle \left(\operatorname{pv}\frac{1}{x}\right)'', \varphi \rangle = -\langle \left(\operatorname{pv}\frac{1}{x}\right)', \varphi' \rangle = \operatorname{pv}\int \frac{\varphi'(x)-\varphi'(0)}{x^2} \ dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{\varphi'(x)-\varphi'(0)}{x^2} \ dx \\ = \lim_{\epsilon\to 0} \left( \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{-\infty}^{-\epsilon} + \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{\epsilon}^{\infty} \\ + 2 \int_{|x|>\epsilon} \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^3} \ dx \right) \\ = 2\operatorname{pv}\int \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^3} \ dx.$$
The question is why the below lim is equal to 0? I think it has to do with the space where generalized functions are defined. I tried brute force but it didn't work because of the uncertainties that arise.
$$\lim_{\epsilon\to 0} \left( \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{-\infty}^{-\epsilon} + \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{\epsilon}^{\infty} \right) \\$$
Any help would be great!
In the theory of distributions, $\phi$ is smooth and of compact suport. And the limit is zero precisely since $\phi\in C_C^\infty$.
Using Taylor's theorem with the Peano form of the remainder, we have as $x\to 0$
$$\phi(x)-\phi'(0)x-\phi(0)=o(x^2)$$
where $o(1)\to 0$ as $x\to 0$.
Moreover, we have as $x\to \pm\infty$
$$\phi(x)-\phi'(0)x-\phi(0)=O(x)$$
Can you finish now?