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?
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 ...... $\displaystyle a_2: y= -x+1,\ x < 0$
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 $