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Math Help - [SOLVED] Finding a limit using l'Hopital's Rule

  1. #1
    Member sinewave85's Avatar
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    [SOLVED] Finding a limit using l'Hopital's Rule

    I thought I had this one, but my answer does not agree with what I got when I checked my answer against a calculator graph of the equation. I got lim = 1, the graph shows lim = 0.

    L = \lim _{x\rightarrow0}\left(\cot{x} - \frac{1}{x}\right)

    L =  \lim_{x\rightarrow0}\left(\frac{1}{\tan{x}} - \frac{1}{x}\right)

    \ln{L} =  \ln{\lim_{x\rightarrow0}\left(\frac{1}{\tan{x}} - \frac{1}{x}\right)}

    \ln{L} = \lim_{x\rightarrow0}\ln\left(\frac{\frac{1}{\tan{x  }}}{\frac{1}{x}}\right)

    \ln{L} = \lim_{x\rightarrow0}\ln\left(\frac{x}{\tan{x}}\rig  ht)

    \ln{L} = \ln\lim_{x\rightarrow0}\left(\frac{x}{\tan{x}}\rig  ht)  \    \mbox{Here is the 0/0 form for l'Hopital's rule.}

    \ln{L} = \ln\lim_{x\rightarrow0}\left(\frac{1}{\sec{x}^2}\r  ight)

    \ln{L} = \ln{1}

    L = 1

    So, where did I go wrong? I have looked over this a dozen times and can not see the problem. Thanks.
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  2. #2
    Eater of Worlds
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    ln\left(\frac{1}{tan(x)}-\frac{1}{x}\right)\neq ln\left(\frac{\frac{1}{tan(x)}}{\frac{1}{x}}\right  )

    You are thinking of ln\left(\frac{\frac{1}{tan(x)}}{\frac{1}{x}}\right  )=ln(cot(x))-ln(\frac{1}{x})

    This is a common mix up.
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  3. #3
    Member Abu-Khalil's Avatar
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    \lim_{x\to 0}\frac{1}{\tan x}-\frac{1}{x}=\lim_{x\to 0}\frac{x-\tan x}{x\tan x}\overbrace{=}^{\text{L'H}}\lim_{x\to 0}\frac{1-\sec^2 x}{\tan x+x\sec^2 x} \overbrace{=}^{\text{L'H}}\lim_{x\to 0}\frac{2\sec^2 x\tan x}{2\sec^2 x+2x\sec ^2 x \tan x}\to 0
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  4. #4
    Member sinewave85's Avatar
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    Thanks for the help! I will have to watch the ln properties -- I am a little rusty. Abu-Khalil's way works better, anyway. I wish I gone that route to begin with -- it would have saved me a lot of time and frustration.
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