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How to find the vertical assimptote

  1. Feb 15, 2009 #1
    [tex]
    f(x)=\frac{1}{1+ln|x|}\\
    [/tex]
    [tex]
    x\neq \frac{1}{e}\\
    [/tex]
    [tex]
    x\neq \frac{-1}{e}\\
    [/tex]
    [tex]
    lim_{x->\frac{1}{e}^+}\frac{1}{1+ln|x|}=\\
    [/tex]
    [tex]
    lim_{x->\frac{1}{e}^-}\frac{1}{1+ln|x|}=\\
    [/tex]
    i cant immagine ithe values in this function
    ??
     
  2. jcsd
  3. Feb 15, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    First, 1/e is a postive number so you can just drop the absolute value.

    Second, ln e-1= -1. Since ln(x) is an increasing function, if x< e-1, ln(x)< -1 and if x> e-1, ln(x)> -1. That's all you need.
     
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