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

Question

1057 views around the world

You can reuse this answer
Creative Commons License

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]}