How shall we show that this limit exists?

[Mike400]

Let:

$\displaystyle f=\int_{V'} \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3}\ dV'$

where $V'$ is a finite volume in space
$\mathbf{r}=(x,y,z)$ are coordinates of all space
$\mathbf{r'}=(x',y',z')$ are coordinates of $V'$
$|\mathbf{r}-\mathbf{r'}|=[(x-x')^2+(y-y')^2+(z-z')^2]^{1/2}$

How to prove that:

$\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x,y,z)-f(x,y,z)}{\Delta x}$ exist

 

[DrClaude]

What have you attempted to find the solution?

 

[Mike400]

What have you attempted to find the solution?

$\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x,y,z)-f(x,y,z)}{\Delta x}\\ =\lim\limits_{\Delta x \to 0}\dfrac{\displaystyle\int_{V'} \dfrac{(x+\Delta x)-x'}{|\mathbf{r}(x+\Delta x,y,z)-\mathbf{r'}|^3}\ dV' - \int_{V'} \dfrac{x-x'}{|\mathbf{r}(x,y,z)-\mathbf{r'}|^3}\ dV'}{\Delta x}\\ =\lim\limits_{\Delta x \to 0}\displaystyle\int_{V'} \dfrac{\left( \dfrac{(x+\Delta x)-x'}{|\mathbf{r}(x+\Delta x,y,z)-\mathbf{r'}|^3} -\dfrac{x-x'}{|\mathbf{r}(x,y,z)-\mathbf{r'}|^3} \right)}{\Delta x}dV'$

Now if only I could take the limit inside the integral (with respect to $V'$), I can proceed to show the limit exists. One of my colleagues told me that we cannot always take such limits inside the integral. This is where I am stuck. Hope someone here can help. Thanks in advance.