How do i find the horizontal asymptote of f(x)=x/(x-1)^2?

Answer 1

You have a vertical asymptote at x=1. For the horizontal
one, set y=1/x in f(x), you get:

f(x) = y/(1-y)^2 = y + 2 y^2 +.... when y ->0

So when x-> +/- Infinity you have:

f(x) = 1/x + order(1/x^2) it goes to 0 as x-> Infinity

Hence the x-axis is the horizontal asymptote at x-> +/- infinity.

The leading term in the expansion above is 1/x, so
at x-> + infinity, the graph of f is above the asymptote.
at x-> - infinity, the graph of f is below the asymptote.
At

Answer 2

if f(x) --> a, where a is finite, when x --> +- infinity then horizontal line y = a is a horizontal asymptote,

x/(x - 1)^2
= x/(x^2 - 2x + 1)
= (x/x^2)/((x^2 - 2x + 1)/x^2) [divide top & bottom by x^2]
= (1/x)/(1 - 2/x + 1/x^2)
--> 0/(1 - 0 + 0) = 0 as x --> +- infinity,
so horizontal asymptote is x-axis.

Answer 3

As x-> ∞ , f(x)->x / x² = 1 / x

f(x) = 1 / x and as x->∞ , f(x) -> 0
i.e. x axis is horizontal asymptote