Undergrad Are Extra Conditions Affecting the Limit Definition?

Click For Summary
The discussion centers on whether additional conditions in the epsilon-delta definition of limits affect the conclusion about the limit of a potential function at a specific point. It is established that by selecting appropriate volumes and intervals, one can make epsilon arbitrarily small, supporting the limit's existence. However, the presence of highlighted terms raises questions about their impact on the limit's definition. The author expresses concern over the choice of symbols for volume and interval, suggesting they may not be optimal. Ultimately, the conversation explores the implications of these conditions on the validity of the limit statement.
Beelzedad
Messages
24
Reaction score
3
TL;DR
Do the following extra highlighted words in the ##\epsilon-\delta## definition of limit prevent us from concluding that the limit exists? Why? Why not?
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}##.

246708

_______________________________________________________________________________________________________________________________________________________________________

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 δ 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 volume δ 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?
 
  • Like
Likes berkeman
Physics news on Phys.org
Beelzedad said:
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'##
This seems to be a reasonable conclusion to me.
However, the author's choice of ##\delta## to represent a volume seems very ill-chosen to me, as well as ##\delta x## for an interval around ##\Delta x = 0##.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 24 ·
Replies
24
Views
4K