How do you find the vertical, horizontal or slant asymptotes for $f(x)=x/(x-1)^2$ ?

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Answer 1 · 2017-01-26T06:25:17

The vertical asymptote is $x=1$
The horizontal asymptote is $y=0$
No slant asymptote

As you cannot divide by $0$ , $x!=1$

The vertical asymptote is $x=1$

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

$lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^-$

$lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^+$

The horizontal asymptote is $y=0$

graph{(y-x/(x-1)^2)(y)(y-100x+100)=0 [-4.38, 4.39, -2.19, 2.193]}

How do you find the vertical, horizontal or slant asymptotes for f(x)=x/(x-1)^2f(x)=x/(x-1)^2?