Does the horizontal asymptote exist or not?

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1 Answer
Truong-Son N.
Dec 15, 2017

The horizontal asymptote occurs when the graph tries to approach #y = "something"# but never reaches it. The limit as #x# gets large would allow this:

#lim_(x->oo) (x^3 - 2x^2 + 3)/(x-2)#

As #x# gets large, the lower-order terms go away and:

#lim_(x->oo) (x^3 - cancel(2x^2 + 3)^"small")/(x-cancel(2)^"small") = x^3/x = x^2#

So the graph is shaped like #x^2# and has no horizontal asymptote. I superimposed #x^2# on here:

graph{((x^3 - 2x^2 + 3)/(x-2) - y)(x^2 - y) = 0 [-20, 20, -50, 150]}

It does have a vertical asymptote at #x = 2# though, where #y# does not exist.

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