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Limit Question

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the following limit:

    [tex]
    \lim_{x \to 0} (1-\text{cos }x)\text{sin }\frac{1}{x}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    (1-cos x) -> 0 as x -> 0. sin (1/x) oscillates infinitely many times as x -> 0.

    intuition tells me that the limit is 0, but how do i show that?

    some ideas i have are using the fact that |sin(1/x)| =< 1, but i'm not sure.
     
  2. jcsd
  3. Oct 25, 2009 #2

    zcd

    User Avatar

    Try the squeeze theorem with something that converges to zero like [tex]\frac{1-\cos{x}}{x}[/tex].
     
  4. Oct 25, 2009 #3
    i ended up doing this.

    [tex]
    \begin{align*}
    -1 &\leq& \text{sin }\frac{1}{x} &\leq& 1\\
    -(1-\text{cos }x) &\leq& (1-\text{cos }x)(\text{sin }\frac{1}{x}) &\leq& 1- \text{cos }x
    \end{align*}
    [/tex]

    since both of the terms on the side equal 0 at x=0, by the squeeze theorem, the middle term also goes to 0.
     
  5. Oct 25, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's how I would have done it. Well done!
     
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