x>0

$\displaystyle

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?

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- Feb 15th 2009, 04:38 AMtransgalacticthe vertical assimptote of this function..
x>0

$\displaystyle

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? - Feb 15th 2009, 04:49 AMearboth
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:

$\displaystyle a_1: y= x-1,\ x > 0$ ...... and ...... $\displaystyle a_2: y= -x+1,\ x < 0$ - Feb 15th 2009, 05:01 AMtransgalactic
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 - Feb 15th 2009, 05:12 AMearboth
- Feb 15th 2009, 05:15 AMtransgalactic
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? - Feb 15th 2009, 05:36 AMearboth
If

$\displaystyle 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:

$\displaystyle 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 $