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Math Help - how to find the vertical assimptote..

  1. #1
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    how to find the vertical assimptote..

    <br />
f(x)=\frac{1}{1+ln|x|}\\<br />
    <br />
x\neq \frac{1}{e}\\<br />
    <br />
x\neq \frac{-1}{e}\\<br />
    <br />
lim_{x->\frac{1}{e}^+}\frac{1}{1+ln|x|}=\\<br />
    <br />
lim_{x->\frac{1}{e}^-}\frac{1}{1+ln|x|}=\\<br />
    i cant immagine ithe values in this function
    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    <br />
f(x)=\frac{1}{1+ln|x|}\\<br />
    <br />
x\neq \frac{1}{e}\\<br />
    <br />
x\neq \frac{-1}{e}\\<br />
    <br />
lim_{x->\frac{1}{e}^+}\frac{1}{1+ln|x|}=\\<br />
    <br />
lim_{x->\frac{1}{e}^-}\frac{1}{1+ln|x|}=\\<br />
    i cant immagine ithe values in this function
    ??
    You'll get the vertical asymptote if

    1+\ln(|x|)=0~\implies~\ln(|x|)=-1~\implies~|x| = e^{-1}=\frac1e

    Now use the definition of the absolute value and you'll get

    x = \frac1e~\vee~x =-\frac1e

    If -\frac1e<x<\frac1e the denominator is negative and if x<-\frac1e~\vee~x>\frac1e the denominator is positive. The function is symmetric to the y-axis. Thus both "limits" you have written above are positive infinity.
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  3. #3
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    i know that but i need to prove it but the limits that i have written
    can you explain how to imagine the value of the vilimit as x goes to its final value??

    i cant see if it goes ti infinity of minus infinity
    ??
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i know that but i need to prove it but the limits that i have written
    can you explain how to imagine the value of the vilimit as x goes to its final value??

    i cant see if it goes ti infinity of minus infinity
    ??
    \lim_{x->\frac{1}{e}^+} \frac{1}{1+ln|x|}

    Let y =\ln(x) then \lim_{x\to\frac1e ^+} y = -1^+

    Then \lim_{x\to\frac1e ^+} (1+y) = 0^+ and therefore \lim_{x->\frac{1}{e}^+} \frac{1}{1+ln|x|} \rightarrow +\infty
    Attached Thumbnails Attached Thumbnails how to find the vertical assimptote..-gebrabsfkt_mitln.png  
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