# Thread: the vertical assimptote of this function..

1. ## the vertical assimptote of this function..

x>0
$
f(x)=|x-1|+\frac{1}{x}\\
$

for m i get:
m=lim |x-1| +1/x=infinity
x->infinity
so there is no vercital assimptote here?

2. Originally Posted by transgalactic
x>0
$
f(x)=|x-1|+\frac{1}{x}\\
$

for m i get:
m=lim |x-1| +1/x=infinity
x->infinity
so there is no vercital assimptote here?
You get vertical asymptotes at those values of x where the function isn't defined. This happens here at x = 0.

Beside the vertical asymptote this graph has two skewed asymtotes:

$a_1: y= x-1,\ x > 0$ ...... and ...... $a_2: y= -x+1,\ x < 0$

3. i agrre that there is an vertical assimptote here on x=0

is there any horisontal assimptote
i cant get the "m" value for it?

y=mx+n

4. Originally Posted by transgalactic
i agrre that there is an vertical assimptote here on x=0

is there any horisontal assimptote
i cant get the "m" value for it?

y=mx+n
No, there aren't any horizontal asymptotes (if by "horizontal" you mean parallel to the x-axis) but there are 2 skewed asymptotes with the slopes m = 1 or m = -1. (Compare my previous post)

5. there is a horisontal assmiptote y=mx+n

m=lim f(x)/x x->infinty
n=lim [f(x)-mx0] x->infinty

and my m limit is not defined so there is no horisontal assimptote
correct?

6. Originally Posted by transgalactic
there is a horisontal assmiptote y=mx+n

m=lim f(x)/x x->infinty No
n=lim [f(x)-mx0] x->infinty

and my m limit is not defined so there is no horisontal assimptote
correct?
If

$m=\lim_{|x|\to \infty}\left(\dfrac{f(x)}x\right)=1,\ x > 0 ~\vee~ m=\lim_{|x|\to \infty}\left(\dfrac{f(x)}x\right)=-1,\ x < 0$

and:

$n=\lim_{|x|\to \infty}\left(f(x)- m\cdot x\right)=-1,\ x > 0 ~\vee~ n=\lim_{|x|\to \infty}\left(f(x)-m\cdot x\right)=1,\ x < 0$