real analysis - Prove the following limit exists

I'm trying understand why $\lim\limits_{\varepsilon \rightarrow 0} \left( \int_{|x| \geq \varepsilon} \frac{\varphi(x)}{x^2} dx - 2 \frac{\varphi(0)}{\varepsilon} \right)$ exists, where $\varphi \in C_c^{\infty}(\mathbb{R})$. The motivation for my question is a linear functional which is defined on the space of distributions by this limit, where $\varphi$ is the input of the linear functional. I would be able to show that the limit of the integral exists, but $\lim\limits_{\varepsilon \rightarrow 0} \frac{\varphi(0)}{\varepsilon}$ not necessarily exist for every $\varphi \in C_c^{\infty}(\mathbb{R})$, then I need compute the whole limit instead of compute the limit by the pieces, but I don't have idea how to do this. I would like to understand why $\lim\limits_{\varepsilon \rightarrow 0} \left( \int_{|x| \geq \varepsilon} \frac{\varphi(x)}{x^2} dx - 2 \frac{\varphi(0)}{\varepsilon} \right)$ exists.

Answer

The Maclaurin series of $\varphi$ begins $a+bx+cx^2+\cdots$. If $a=b=0$,
of course there is no problem. We'd like to reduce to this case by subtracting off

$a+bx$, but that does not have compact support. But we can get around that by
taking a compactly supported and smooth $\psi$ which is equal to $1$ on the
interval $[-1,1]$. Then

$$\varphi=\varphi_1+\varphi_2$$
where $\varphi_1(x)=(a+bx)\psi(x)$
and $\varphi_2\sim cx^2$ as $x\to0$. So we can reduce to considering $\varphi_1(x)$.

For $0<\epsilon<1$,

$$\begin{align} \int_{|x|\ge\epsilon}\frac{\varphi_1(x)}{x^2}\,dx &=\int_{|x\ge1}\frac{\varphi_1(x)}{x^2}\,dx+\int_{\epsilon}^1\frac{a+bx}{x^2}\,dx +\int_{-1}^{-\epsilon}\frac{a+bx}{x^2}\,dx\\ &=C+\int_{\epsilon}^1\frac{2a}{x^2}\,dx=C'+\frac{2\varphi_1(0)}{\epsilon} \end{align}$$
where $C$ and $C'$ are independent of $\epsilon$. In this case, the limit
is just $C'$.