# Thread: how to find the vertical assimptote..

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

$
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
??

2. 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 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.

3. 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
??

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