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Homework Help: I need some help with a proof

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data

    the problem is: prove that if lim x-->0 g(x)=0, then lim x-->0 g(x)sin(1/x)=0. it seems simple enough, but certain parts have me wondering.

    2. Relevant equations

    so i already know that lim x-->0 sin(1/x) is undefined because the function oscillates near 0. i'm not really sure how to incorporate that without just stating it, though. also, how do i show the effect of lim x-->0 g(x)=0? if a limit is undefined, does it not exist?

    3. The attempt at a solution

    i'm not sure if this is completely erroneous but my tentative solution is the following:
    if lim x-->0 g(x)=0, lim x-->0 g(x)sin(1/x) = [lim x-->0 g(x)]*[lim x-->0 sin(1/x)]= 0*lim x-->0 sin(1/x)
    since a number times 0 equals 0, it follows that lim x-->0 g(x)sin(1/x) = 0.

    please help!
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2
    This argument is erroneous since [itex]\lim_{x\rightarrow 0} \sin \frac{1}{x}[/itex] is not a number, and furthermore the limit of the product is not the product of the limits of the factors unless the limits of the factors both exist.
    Since the second limit does not exist, you will have to perform an epsilon-delta proof of the statement, or use a theorem about limits of bounded functions.
  4. Sep 19, 2010 #3
    would i start off showing the epsilon-delta proof of lim x-->0 g(x)=0, or of lim x-->0 g(x)sin(1/x) = 0? since lim x-->0 sin(1/x) DNE, it doesnt have an applicable epsilon-delta form, so how can i include it in an epsilon-delta proof?
  5. Sep 19, 2010 #4
    Have you tried the squeeze theorem?
  6. Sep 19, 2010 #5
    how would squeeze theorem apply? i don't know if g(x)[tex]\leq[/tex]g(x)sin(1/x) or any of the other conditions for squeeze theorem.

    ah, i see now. squeeze theorem can be used because i know the range of the sine function. that definitely clears things up, as i already proved the theorem earlier. thanks for the help!
    Last edited: Sep 19, 2010
  7. Sep 19, 2010 #6
    The sine function is bounded.
    [tex] -1 \leq sin(1/x) \leq 1[/tex]

    Now incorporate g(x).
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