how to find the vertical assimptote..

**transgalactic** · February 15th 2009, 04:52 AM

$ f(x)=\frac{1}{1+ln|x|}\\ $
$ x\neq \frac{1}{e}\\ $
$ x\neq \frac{-1}{e}\\ $
$ lim_{x->\frac{1}{e}^+}\frac{1}{1+ln|x|}=\\ $
$ lim_{x->\frac{1}{e}^-}\frac{1}{1+ln|x|}=\\ $
i cant immagine ithe values in this function
??

**earboth** · February 15th 2009, 05:08 AM

Originally Posted by transgalactic

$ f(x)=\frac{1}{1+ln|x|}\\ $
$ x\neq \frac{1}{e}\\ $
$ x\neq \frac{-1}{e}\\ $
$ lim_{x->\frac{1}{e}^+}\frac{1}{1+ln|x|}=\\ $
$ lim_{x->\frac{1}{e}^-}\frac{1}{1+ln|x|}=\\ $
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.

**transgalactic** · February 15th 2009, 05:11 AM

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
??

**earboth** · February 15th 2009, 05:45 AM

Originally Posted by transgalactic

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 $\lim_{x->\frac{1}{e}^+} \frac{1}{1+ln|x|} \rightarrow +\infty$

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