This question consists of two parts: preliminary and the main question. Reading only the main question may be enough to get my point, but if you want details please have a look at the preliminary.
PRELIMINARY:
Let potential due to a small volume $\delta$ at a point $(1,2,3)$ inside it be denoted by $\psi_{\delta}$.
It can be shown that for every $\epsilon>0$, we can choose a volume $\delta$ such that:
$\left| \dfrac{\psi_{\delta}(1+\Delta x,2,3)-\psi_{\delta}(1,2,3)}{\Delta x} \right| < \dfrac{\epsilon}{3} \tag1$
That is $\epsilon$ can be made as small as we can by choosing a small volume $\delta$.
Also, using spherical coordinate system we can show that for every $\epsilon>0$, we can choose a volume $\delta$ such that:
$\displaystyle\left| \iiint_{\delta} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV' \right| < \dfrac{\epsilon}{3}$
That is:
$\displaystyle\left| \iiint_{V'} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV' -\iiint_{(V'-\delta)} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV' \right| < \dfrac{\epsilon}{3} \tag2$
That is $\epsilon$ can be made as small as we can by choosing a small volume $\delta$.
No matter what our volume $\delta$ is, at point $P(1,2,3)$, $\dfrac{\partial \psi_{(V'-\delta)}}{\partial x}$ exists (since $P$ being an outside point of $V'-\delta$). That is:
$\lim\limits_{\Delta x \to 0} \dfrac{\psi_{(V'-\delta)}(1+\Delta x,2,3)-\psi_{(V'-\delta)}(1,2,3)}{\Delta x}=\dfrac{\partial \psi_{(V'-\delta)}}{\partial x} (1,2,3)$
That is, for every $\epsilon>0$, we can choose an interval $\delta x$ around $\Delta x = 0$ (inside volume $\delta$) such that whenever $0 < |\Delta x-0| < \delta x$:
$\left| \dfrac{\psi_{(V'-\delta)}(1+\Delta x,2,3)-\psi_{(V'-\delta)}(1,2,3)}{\Delta x} - \dfrac{\partial \psi_{(V'-\delta)}}{\partial x} (1,2,3) \right| < \dfrac{\epsilon}{3}$
That is:
$\left| \dfrac{\psi_{(V'-\delta)}(1+\Delta x,2,3)-\psi_{(V'-\delta)}(1,2,3)}{\Delta x} - \left( -\displaystyle\iiint_{(V'-\delta)} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV' \right) \right| < \dfrac{\epsilon}{3} \tag3$
That is $\epsilon$ can be made as small as we can by choosing a small interval $\delta x$ around $\Delta x = 0$ (inside volume $\delta$)
Adding inequalities $(1), (2)$ and $(3)$:
$\left| \dfrac{\psi_{V'}(1+\Delta x,2,3)-\psi_{V'}(1,2,3)}{\Delta x} - \left( -\displaystyle \iiint_{V'} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV' \right) \right| < \epsilon$
That is, for every $\epsilon>0$, we can choose $\bbox[yellow]{\text{a volume $\delta$ and}}$ an interval $\delta x$ around $\Delta x = 0$ (inside volume $\delta$) such that whenever $0 < |\Delta x-0| < \delta x$, the above inequality holds.
That is, $\epsilon$ can be made as small as we can by choosing $\bbox[yellow]{\text{a small volume $\delta$ and}}$ a small interval $\delta x$ around $\Delta x = 0$ (inside volume $\delta$)
QUESTION:
Since there are some extra highlighted words in the above $\epsilon-\delta$ definition of limit, will this prevent us from saying that:
$\lim\limits_{\Delta x \to 0} \dfrac{\psi_{V'}(1+\Delta x,2,3)-\psi_{V'}(1,2,3)}{\Delta x}=-\displaystyle \iiint_{V'} \dfrac{\rho'}{R^2} \dfrac{x-x'}{R} dV'$
? Why? Why not?