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Math Help - Limit Problem (L'Hopital's Rule)

  1. #1
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    Limit Problem (L'Hopital's Rule)

    Edit: I Think the equation came out wrong in latex. Here is the updated one.

    34. lim_{x\to\infty}\frac{sin^2x - 2x}{x^2sin^2x}

    I have done it out, and for the numerator I get 2cos^2x - 2sin^2x -2 and I just know it wants to be a pythagorean and cancel out leaving me with 0. I think I am messing up somewhere because I took it to the second derivation and it just gets funny after that.

    Any help would be appreciated.
    Last edited by Warrenx; December 5th 2010 at 01:51 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The limit does not exist.

    (i) \displaystyle\lim_{x \to{+}\infty}{(1/x^2)}=0

    However, choosing:

    x_n=\dfrac{\pi}{2}+2n\pi,\;y_n=\dfrac{\pi}{4}+2n\p  i\;\;(n\in\mathbb{N})

    we have:

    (ii) \displaystyle\lim_{n \to{+}\infty}{\csc^2 x_n}=1,\;\displaystyle\lim_{n \to{+}\infty}{\csc^2 y_n}=2

    Regards.

    Fernando Revilla
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