Can we prove the limit of sin(x)/x is 1 using the ε-δ definition?

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Discussion Overview

The discussion revolves around the possibility of proving that the limit of sin(x)/x as x approaches 0 equals 1, specifically using the ε-δ definition of limits. Participants explore various approaches, including the Squeeze Theorem, derivatives, and alternative proofs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests a possible proof using the ε-δ definition to show that lim_{x→0} sin(x)/x = 1.
  • Another participant references the Squeeze Theorem as a potential method but expresses skepticism about its applicability due to the non-existence of lim_{x→0} ±1/x.
  • Some participants propose that standard proofs might involve geometric arguments, Taylor series, or L'Hôpital's theorem.
  • One participant mentions using derivatives, stating that taking the derivative of sin(x) at x = 0 yields 1, which they consider sufficient for a physics forum.
  • Another participant points out that using the derivative to prove the limit requires prior knowledge that d/dx sin(x) = cos(x), which introduces circular reasoning unless proven separately.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the Squeeze Theorem for this limit and the appropriateness of using derivatives as a proof method. The discussion remains unresolved with multiple competing perspectives on how to approach the proof.

Contextual Notes

Some limitations include the dependence on the validity of the Squeeze Theorem and the assumptions required for using derivatives in the proof. The discussion does not resolve these issues.

evagelos
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possible proof??

Can we prove that the limit of :

[tex]lim_{n\to 0}\frac{sinx}{x} =1[/tex]

By using the ε-δ definition??
 
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I don't believe the Squeeze Theorem works too well for this one, because [tex]\lim_{ x \to 0 } \pm \frac{1}{x}[/tex] does not exist. The standard proof of this fact would cite either a geometric argument, Taylor series, or L'Hopital's theorem.
 


i dunno, take the derivative and set x equal to 0 and you get 1. that's proof enough for a physics forum
 


d/dx sin(x) @ x = 0, is defined as: lim x->0 [sin(x) - sin(0)]/x = lim x->0 sin(x)/x.

Basically, you can't assume that this limit is 1 to prove that this limit is 1, unless ofcourse you can prove that d/dx sin(x) = cos(x) and consider this as a special case (however, you will come to find that hidden in this proof contains the question being asked).
 

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