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Thread: L'Hopital's Rule

  1. #1
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    L'Hopital's Rule

    Please prove
    lim x-> 0+ (ln cotx)/(e^csc(^2)x) = 0 using L'Hopital's Rule.

    Thanks!
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  2. #2
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    Quote Originally Posted by ricey View Post
    Please prove
    lim x-> 0+ (ln cotx)/(e^csc(^2)x) = 0 using L'Hopital's Rule.

    Thanks!
    This is tough, but not impossible.

    We first need to check that this is of indeterminate form.

    It is, since the top $\displaystyle \to \infty$ and the bottom $\displaystyle \to \infty$.


    So we need to take the derivative of the top and the derivative of the bottom.

    $\displaystyle \frac{d}{dx}[\ln{(\cot{x})}] = -\frac{1}{\sin{x}\cos{x}}$


    $\displaystyle \frac{d}{dx}\left[e^{\csc^2{x}}\right] = -\frac{2\cos{x}e^{\csc^2{x}}}{\sin^3{x}}$.


    So $\displaystyle \lim_{x \to 0^+}\frac{\ln{(\cot{x})}}{e^{\csc^2{x}}} = \lim_{x \to 0^+}\frac{-\frac{1}{\sin{x}\cos{x}}}{-\frac{2\cos{x}e^{\csc^2{x}}}{\sin^3{x}}}$

    $\displaystyle = \lim_{x \to 0^+}\frac{\tan^2{x}}{2e^{\csc^2{x}}}$


    As $\displaystyle x \to 0, \csc^2{x} \to \infty$.

    So the bottom $\displaystyle \to \infty$.


    Therefore this whole limit tends to $\displaystyle \frac{0}{\infty} = 0$.
    Last edited by Prove It; Jul 28th 2009 at 05:44 PM.
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  3. #3
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    would deriv of ln cot x be sinx/[(cosx )(-sin ^(2)x]

    = 1/ [(-cosx) (sinx)]
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  4. #4
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    Quote Originally Posted by ricey View Post
    would deriv of ln cot x be sinx/[(cosx )(-sin ^(2)x]

    = 1/ [(-cosx) (sinx)]
    Yes it is.

    I'll fix it now...
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  5. #5
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    okay phew lol
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