Precise definition of a limit

  • Thread starter Shaybay92
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  • #1
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I would really appreciate if someone could please explain to me how the final result of the proof of a limit actually PROVES the limit? http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html

I was particularly looking at Solution 2 on this website. I understand how |x-10| was factorized out etc. and therefore it can equal delta... but when it says that if delta = epsilon/3, |x-10| < epsilon/3 and therefore |f(x)-35| is less than epsilon.... how is this true??? I mean we dont even know what epsilon is?

how does this even prove that the limit is 35
 
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Answers and Replies

  • #2
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I understand how |x-10| was factorized out etc. and therefore it can equal delta

Does it?

Do you understand the epsilon-delta argument. It is a two part formal method much used in mathematical analysis. Would this be your difficulty?
 
  • #3
tiny-tim
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Hi Shaybay92! :smile:

(have a delta: δ and an epsilon: ε :wink:)
… I mean we dont even know what epsilon is?

ε is anything (> 0).

Continuity is proved if, for any ε, you can find a δ (which depends on that ε) which works. :wink:
 
  • #4
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ok but how does this proof tell us that there is a delta for any epsilon? i dont really see how if delta = cepsilon (c is some constant) then that is a proof...
 
  • #5
tiny-tim
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Hi Shaybay92! :smile:

(what happened to that δ and ε i gave you? :wink:)
ok but how does this proof tell us that there is a delta for any epsilon? i dont really see how if delta = cepsilon (c is some constant) then that is a proof...

because, if δ is defined as cε, then for any ε, there is a δ ! :smile:
 
  • #6
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so as long as x is less than a particular multiple of ε (cε) then |f(x)-L| < ε?
 
  • #7
tiny-tim
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so as long as x is less than a particular multiple of ε (cε) then |f(x)-L| < ε?

That's right. :smile:

(Warning: although δ is usually a multiple of ε, somtimes it's something more awkward, like ε2 :wink:)
 

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