# How to find the vertical assimptote

1. Feb 15, 2009

### 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
??

2. Feb 15, 2009

### HallsofIvy

Staff Emeritus
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.