Undergrad Proving Theorem 1 in Spivak's Calculus: Tips & Tricks

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SUMMARY

This discussion focuses on proving Theorem 1 from Spivak's "Calculus," which states that a function cannot approach two different limits near a point. The key points of confusion involve the choices of delta and epsilon in the proof. Specifically, delta is chosen as the minimum of two distances (min(d1, d2)), and epsilon is set to half the difference between the two limits (|L - M| / 2). Understanding these choices clarifies the proof structure and reinforces the geometric interpretation of limits.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with epsilon-delta definitions of limits
  • Basic knowledge of proof techniques, particularly proof by contradiction
  • Ability to interpret geometric representations of functions
NEXT STEPS
  • Study the epsilon-delta definition of limits in detail
  • Explore proof techniques in calculus, focusing on proof by contradiction
  • Review geometric interpretations of limits and continuity
  • Practice problems involving the application of limits with varying delta and epsilon values
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Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limit proofs and epsilon-delta definitions in mathematical analysis.

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Hello

I am struggling with proving theorem 1, pages 98-99, in Spivak's Calculus book: "A function f cannot approach two different limits near a."

I understand the fact that this theorem is correct. I can easily convince myself by drawing a function in a coordinate system and trying to find two different limits at a given x coordinate and it will not work.

But when proving the theorem I fail to see the notion behind two choices that Spivak made in proving this theorem:

(i) he chooses delta = min(d1, d2), and
(ii) he chooses epsilon = |L - M| / 2

I understand the structure of the proof, which is a proof by contradicting the assumption that L unequal M. But I am stuck at the above two mentioned choices of delta and epsilon.

I apologize sincerely for not using Latex symbols and notation and for not posting pictures of the text, but atm I am on my smartphone and do not have access to a computer.

Any reference/help is appreciated!
 
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I downloaded the PF app. And I noticed that one can use it to upload pictures.

The problem and part 1 of the proof:

IMG_1473781452.714081.jpg


Part 2 of the proof:

IMG_1473781498.193049.jpg
 
In terms of choosing epsilon to be half the difference between the limits:

Eventually the function must get within epsilon of both limits. But it can't be less than half the distance from them both at the same time.

if you and a friend stand 1m apart. No one can stand within 0.5m of you both at the same time.
 
Now that you put it in terms of distance it makes sense. But how does one develop the intuition to "see" what value for a variable one should choose when proving theorems?

I figured the delta part out:

Since the definition of a limit states, that "for all epsilon > 0, there is some delta > 0,..." It means that if we have the two deltas above mentioned one can always choose a smaller one, thus by taking the min(d1, d2) makes also sense.

Thank you for your reply!
 
I guess a lot of people see the epsilon- delta definition as mysterious, but it always seemed to me a fairly logical way of formalising the geometric behavior of a continuous function.

Take a fresh look at it from that perspective perhaps.
 
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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