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Math Help - Having trouble taking this seemingly simple limit..

  1. #1
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    Having trouble taking this seemingly simple limit..

    I'm having trouble taking this limit which should definitely be really simple. It looks as follows:

    limit as x goes to 1 of (1 - x + lnx)/(1 + cos pi x)

    This should probably use l'hopital's rule, since it goes to 0/0 so I take the derivative and get (-1 + 1/x)/(-sin pi x) which equals (-1 + x^-1)/(-sin pi x). This is 0/0 again, so I take l'hopital's rule again. I get (x^-2)/cos pi x. This is -1 as x approaches 1. However, I know this is wrong.

    Where have I gone wrong?

    Also, how are math equations done on this forum properly? Does it use TeX?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by charleschafsky View Post
    I'm having trouble taking this limit which should definitely be really simple. It looks as follows:

    limit as x goes to 1 of (1 - x + lnx)/(1 + cos pi x)

    This should probably use l'hopital's rule, since it goes to 0/0 so I take the derivative and get (-1 + 1/x)/(-sin pi x) which equals (-1 + x^-1)/(-sin pi x). This is 0/0 again, so I take l'hopital's rule again. I get (x^-2)/cos pi x. This is -1 as x approaches 1. However, I know this is wrong.

    Where have I gone wrong?

    Also, how are math equations done on this forum properly? Does it use TeX?
    When you differentiate cos \pi x you get -\pi sin \pi x. Differentiating -\pi sin \pi x gives -\pi^2 cos \pi x.

    This forum uses latex for typing math formulas. Quote my post to learn to use it.
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  3. #3
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    Hello, charleschafsky!

    \lim_{x\to1}\frac{1 - x + \ln x}{1 + \cos\pi x}

    This should use L'hopital's rule, since it goes to \tfrac{0}{0}.

    So I take the derivative and get: . \frac{-1 + \frac{1}{x}} {-{\color{red}\pi}\sin \pi x} . . . . Chain rule!
    which equals: . \frac{-1 + x^{-1}}{-\pi\sin\pi x}

    This is \tfrac{0}{0} again, so I use L'hopital's rule again.

    I get: . \frac{x^{-2}}{{\color{red}\pi^2}\cos\pi x}

    Therefore: . \lim_{x\to1}\frac{\frac{1}{x^2}}{\pi^2\cos\pi x} \;=\;\frac{\frac{1}{1^2}}{\pi^2\cos\pi} \;=\;\frac{1}{\pi^2(-1)} \;=\;-\frac{1}{\pi^2}

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