Undergrad Are Extra Conditions Affecting the Limit Definition?

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SUMMARY

The discussion centers on the implications of extra conditions in the limit definition of a function involving potential, specifically at the point (1,2,3). It establishes that for any ε>0, a volume δ can be chosen such that the limit definition holds true, allowing the derivative to exist. The primary question raised is whether the presence of additional highlighted terms in the ε-δ definition affects the validity of the limit statement. The consensus is that these extra conditions do not prevent the limit from being defined as stated, although the choice of symbols δ and δx for volume and interval respectively is criticized.

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  • Understanding of ε-δ definitions in calculus
  • Familiarity with multivariable calculus concepts
  • Knowledge of spherical coordinate systems
  • Basic principles of limits and derivatives
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  • Study the ε-δ definition of limits in detail
  • Explore multivariable calculus, focusing on partial derivatives
  • Learn about the application of spherical coordinates in calculus
  • Investigate common pitfalls in limit definitions and their implications
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Students and professionals in mathematics, particularly those studying calculus and analysis, as well as educators seeking to clarify concepts related to limits and derivatives.

Beelzedad
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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}##.

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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##.
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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##.
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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##)
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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?
 
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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##.
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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