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Math Help - trig limits

  1. #1
    Member purplec16's Avatar
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    trig limits

    lim x-> 0 x^2/(1-cos2x)

    lim x-> 0 x*(x/(1-cos2x))

    lim x-> 0 (x/2)(2x/1-cos2x)

    lim x-> 0 x/2*0

    =0/2*0

    =0

    just checking to see if i did the right thing
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  2. #2
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    Re: trig limits

    Quote Originally Posted by purplec16 View Post
    lim x-> 0 x^2/(1-cos2x)
    lim x-> 0 x*(x/(1-cos2x))
    lim x-> 0 (x/2)(2x/1-cos2x)
    lim x-> 0 x/2*0
    =0/2*0
    =0
    That is not correct.
    This is ideal for using l'Hopiial's rule twice.
    If you don't have the use of l'HR then note that
    \frac{{x^2 }}{{1 - \cos (2x)}} = \frac{1}{2}\left( {\frac{x}{{\sin (x)}}} \right)^2 .
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  3. #3
    Member purplec16's Avatar
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    Re: trig limits

    No we didnt learn L'Hospitals Rule
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  4. #4
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    Re: trig limits

    when you have a limit of the form 0/0 (as x--> a), you can't just call it "0" (such a limit is called "indeterminate").

    hopefully, you have already covered what lim x-->0 sin(x)/x is.
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