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Math Help - L'Hopitals Rule

  1. #1
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    L'Hopitals Rule

    I need to prove the following limit exists, and evaluate it

    lim
    x-->0 1-cos(3x)/x^4+3x^2

    Let f(x) = 1-cos(3x) and g(x) = x^4+3x^2

    The f and g are differentiatiable, and f(0)=g(0)=0

    Thus f and g satisfy the conditions of L'Hopitals Rule at 0.

    Now

    f'(x)=sin3x and g'(x)= 4x^3+6x

    Thus by L'Hopitals Rule, the required limit exists and equals

    lim
    x-->0 f'(x)/g'(x) = lim x-->0 Sin3x/4x^3+6x = ?

    This is where I am stuck. I know if just the basic end bit but could anyone help?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: L'Hopitals Rule

    Quote Originally Posted by Arron View Post
    I need to prove the following limit exists, and evaluate it

    lim
    x-->0 1-cos(3x)/x^4+3x^2

    Let f(x) = 1-cos(3x) and g(x) = x^4+3x^2

    The f and g are differentiatiable, and f(0)=g(0)=0

    Thus f and g satisfy the conditions of L'Hopitals Rule at 0.

    Now

    f'(x)=sin3x and g'(x)= 4x^3+6x

    Thus by L'Hopitals Rule, the required limit exists and equals

    lim
    x-->0 f'(x)/g'(x) = lim x-->0 Sin3x/4x^3+6x = ?

    This is where I am stuck. I know if just the basic end bit but could anyone help?
    First a little observation: if f(x)= 1-\cos 3x then f^{'}(x) = 3\ \sin 3x...

    Second: if the first application of l'Hopital rule gives an indeterminate form, then try to use l'Hopital rule again till to obtain a non indeterminate form!...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: L'Hopitals Rule

    Thanks

    So f"(x)= 9 cos 3x
    and g"(x) = 12x^2+6

    Since g"(0) = 6 which is not 0. we deduce, by combination and continuous functions, that

    lim
    x---> =9/6 = 3/2

    Is this correct?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: L'Hopitals Rule

    Correct!
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