How shall we show that this limit exists?

  • #1
58
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Moved from a technical forum
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
 

Answers and Replies

  • #2
DrClaude
Mentor
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What have you attempted to find the solution?
 
  • #3
58
6
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.
 

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