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Math Help - Question regarding limits. Help appreciated!

  1. #1
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    Question regarding limits. Help appreciated!

    what is lim(x,y) approaching zero sinxcos(1/y) by using squeeze theorem?
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Question regarding limits. Help appreciated!

     -1 \leq cos(\frac{1}{y}) \leq 1 and so
    tiny condondrum to the above as sin(x) can take on negative values.

    So more apt would be either
     -sin(x) \leq sin(x) cos(\frac{1}{y}) \leq sin(x)
    or
     -sin(x) \geq sin(x) cos(\frac{1}{y}) \geq sin(x)

    either way the limit is the same for sin(x) = 0  x \to 0
    Last edited by jakncoke; February 12th 2013 at 11:36 AM.
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