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Limits - Prove that xsin1/x approaches 0 near 0

  1. May 25, 2012 #1
    limits -- Prove that xsin1/x approaches 0 near 0

    Prove that xsin1/x approaches 0 near 0.

    Similiar Proof from book
    |sin1/x| ≤ 1
    | xsin1/x | ≤ |x| for all x not equal to 0, so we can make |xsin1/x|< ε by requiring that |x| < ε and not equal to 0.

    MY QUESTION: Prove x2sin1/x approaches 0 near 0.

    According to the book, if ε>0, to ensure that |x2sin1/x |< ε we need only require that |x| < ε and not equal to 0.

    shouldnt |x| be less that √ε? Hence |x2 | < ε and |x|< √ε.
     
  2. jcsd
  3. May 25, 2012 #2

    sharks

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    Use the Pinching theorem.
    [tex]-x \le x\sin \frac{1}{x} \le x[/tex]
     
  4. May 25, 2012 #3
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    Ok how does that tell me that |x2sin1/x|< ε for |x|< ε??
     
  5. May 25, 2012 #4

    SammyS

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    Are you supposed to use a δ - ε proof, or can you use such things as the "squeeze theorem" ?
     
  6. May 25, 2012 #5
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    delta epsilon proof
     
  7. May 25, 2012 #6
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    all I am doing is quoting from the book... and I want to know how they decided that |x|<ε
     
  8. May 25, 2012 #7
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    Well, for small enough [itex]\epsilon[/itex], [itex]0<\epsilon < \sqrt{\epsilon}[/itex].

    In fact, we only need [itex] 0<\epsilon<1[/itex] for this to be true.

    Oh and also for a more fundamental reason. In Spivak's book, (I don't know for other books) the proof that this limit is 0 using delta-epsilon comes before the proof that every positive number has a square root (which requires the least upper bound property). So the existence of such a number [itex]\sqrt{\epsilon}[/itex]. cannot be assumed.
     
    Last edited: May 25, 2012
  9. May 25, 2012 #8
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    This is a common hangup for people doing ##\epsilon-\delta## proofs for the first time.

    You don't need to find the "best" ##\delta##; you just need to find one that works.
     
  10. May 25, 2012 #9

    SammyS

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    If [itex]\displaystyle |x| < 1\,,\ \text{ then (multiplying by }|x| \text{ gives } |x^2|<|x|\,,\text{ and } |x| < 1\, \text{ ) } |x^2|<1 \ .[/itex]

    So let δ = min(1,√ε) .
     
  11. May 25, 2012 #10
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    There is a general theorem:
    If [itex]f(x) \rightarrow 0, \ x \rightarrow a[/itex], and [itex] \vert g(x) \vert \le M, \ \forall x \in U(a)[/itex], where [itex]U(a)[/itex] is some neighborhood of the point [itex]a[/itex], then:
    [tex]
    \mathrm{\lim}_{x \rightarrow a} f(x) \, g(x) = 0
    [/tex]

    Can you prove it?
     
  12. May 26, 2012 #11
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    I just learned the δε definition myself and am very shaky on it. Since |x|< δ can we not turn |x2sin([itex]\frac{1}{x}[/itex])|< ε into |x|<[itex]\frac{ε}{xsin(\frac{1}{x})}[/itex]. And now we have our δ=[itex]\frac{ε}{xsin(\frac{1}{x})}[/itex] and use this in the proof.


    Because it's part of the εδ definition that |x2sin(1/x)|<ε and it's just the nature of the function that |x| ≤ |xsin(1/x)|
     
    Last edited: May 26, 2012
  13. May 26, 2012 #12

    micromass

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    No, this way our δ would depend on x.

    The ε-δ definition tells us that "For all ε, there must exist a δ, such that for all x, ...". This means that our δ must work for all x. Your proposed δ does not work for all x.
     
  14. May 26, 2012 #13

    HallsofIvy

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    For [itex]0< \epsilon< 1[/itex], [itex]\epsilon< \sqrt{\epsilon}[/itex] so "[itex]<\epsilon[/itex]" is sufficient.
     
  15. May 26, 2012 #14
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    You've got your quantifiers mixed up an have given the definition of uniform continuity.
     
  16. May 26, 2012 #15

    micromass

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    No, I didn't...

    Uniform continuity would be "Forall ε>0, there is a δ>0, such that for all x,y..."
    Here, I'm holding 0 fixed and I say " "Forall ε>0, there is a δ>0, such that for all x..."

    OK, it might be confusing though that I did not write the definition fully.
     
  17. May 26, 2012 #16

    SammyS

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    Re: limits -- Prove that xsin1/x approaches 0 near 0

    I must have had a brain cramp!

    It looks like you could use either of the following for δ :
    δ = √ε

    δ = min(1, ε)​
     
  18. May 26, 2012 #17
    Yes. You are correct. I misread your yada-yada. Though I probably shouldn't have given the context.

    My bad.
     
  19. May 27, 2012 #18
    Re: limits -- Prove that xsin1/x approaches 0 near 0

    If you want to be fancy, the power series expansion of sinx is:

    sinx = x - x3/6 + ....

    It follows that sin(1/x) = 1/x - (1/x)3/6 + ...

    and x2sin(1/x) = x - (1/x)/6 + ...

    so, lim x -> ∞ x2sin(1/x) = 0
     
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