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Finding the limit of a trig function

  1. Sep 20, 2007 #1
    I'm stuck on this limit function and I don't know what to do next. Please help me out. Thanks!

    [tex]\lim_{x\rightarrow 0^{+}}(\frac{\csc2x}{x})[/tex]

    I just turned csc2x into 1\sin2x so then I have:

    [tex] \lim_{x\rightarrow 0^{+}} (\frac{1}{x\sin2x})[/tex]

    Then I used the trig identity: sin2x=2sinxcosx

    ...but I'm not sure if this is going to take me anywhere.

    I know that [tex]\lim_{x\rightarrow c} \sin x = \sin c[/tex] but I'm not sure how to incoporate that identity in this problem.
     
  2. jcsd
  3. Sep 20, 2007 #2
    You should use l'Hopital's rule. Or just note that 1/0 is never going to be a good thing...
     
  4. Sep 20, 2007 #3
    Would the squeeze theorem work? -1<sinxcosx<1 to prove that the limit = 0?
     
  5. Sep 21, 2007 #4

    Gib Z

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    Homework Helper

    The easiest way is genneths second comment. Just realise that the denominator, x sin 2x, is going to 0, and there is nothing on the numerator to cancel out with. Denominator approaches 0, whole thing approach infinity.
     
  6. Sep 21, 2007 #5
    You can t use L'Hospitals rule. It has to be of the form 0/0 or inf/inf. Since this is neither, try to use the identity lim x->0 sinx/x =1.
     
  7. Sep 21, 2007 #6
    To be honest, this is one of those limits which doesn't need a limit. You just put zero in, and go, "oh, it's 1/0". No funky limit taking will change that.
     
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