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Math Help - finding the limit of a trig function

  1. #1
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    finding the limit of a trig function

    I know i got this one wrong somewhere along the way...

    lim x goes to 0 (2x)cot(3x)

    this is what I did:

    lim x goes to 0 ((2xcos3x)/sin3x) = lim x to 0 (2xcos3x)/3x time lim to 0 sin 3x/3x
    = 2 times lim x to 0 cos3x/3x times lim x to 0 sin 3x/3x
    both cose x/x and sinx/x cancel out to one
    therefore: 2(3)(3) = 18

    I think I messed up on the manipulation somewhere...
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  2. #2
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    Re: finding the limit of a trig function

    Quote Originally Posted by tmas View Post
    I know i got this one wrong somewhere along the way...

    lim x goes to 0 (2x)cot(3x)
    Write it as \left( {\frac{{2\cos (3x)}}{3}} \right)\left( {\frac{{3x}}{{\sin (3x)}}} \right)
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