# How to prove the limit exist?

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

$$\text{ }$$

MY TRY:

I am not sure whether this method would work. If it doesn't please suggest another method to reach my goal.

\begin{align} &\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' \end{align}

Now if only I could take the limit inside the integral (with respect to $$V′$$),I can proceed to show the limit exists.

If we can't do that and this method doesn't work, please suggest another method to show that the limit exists.

• You wrote $f(x)$ and $f(x+\Delta x)$ -- what about the $y$ and $z$? – user10354138 May 7 at 15:38
• $y$ and $z$ are constants. Let me edit – Joe May 7 at 16:01
• You want $\int_{V'}\dots$ not $\int_V\dots$ in the first formula so you are only integrating over a finite region. Show that for fixed $x,y,z$, $g(x',y',z')=\frac{\partial}{\partial x}\frac{x-x'}{\lvert\mathbf{r}-\mathbf{r}'\rvert^3}$ is absolutely integrable on $V'$. – user10354138 May 8 at 1:21
• @user10354138 Then what shall we do to reach our conclusion that the limit exists. Can you please elaborate in an answer. – Joe May 8 at 8:24

Assume that $$C$$ is a $$3$$-dimensional body with a smooth boundary and $${\rm vol}\ C<\infty$$. Define $$f(\textbf{r}) =\int_C\ \frac{x-x'}{|\textbf{r}-\textbf{r}'|^3}\ d{\rm vol}\ (\textbf{r}' )$$ Prove that $$f$$ is finite.
Proof : $$\int_{B_\epsilon (0)}\ \frac{1}{|{\bf r}|^2} \ d{\rm vol}\ ({\bf r}) \leq C\epsilon$$ for some $$l>0$$ and note that $$\frac{|x-x'|}{|\textbf{r}-\textbf{r}'|^3}\leq \frac{1}{| \textbf{r}-\textbf{r}'|^2}$$
So $$|f( \textbf{r} )| \leq \int_{B_\epsilon ({\bf r} )}\ \frac{|x-x'|}{|\textbf{r}-\textbf{r}'|^3}\ d{\rm vol}\ (\textbf{r}') +\int_{C-B_\epsilon(\textbf{r}) }\ \frac{1}{|\textbf{r}- \textbf{r}'|^2}\ d{\rm vol}\ (\textbf{r} ')$$
$$\leq l\epsilon +\int_{C-B_\epsilon(\textbf{r}) }\ \frac{1}{\epsilon^2} \ d{\rm vol}\ (\textbf{r} ') \leq l\epsilon + \frac{1}{\epsilon^2}{\rm vol}\ C$$