# Evalute the limit (6x^3-7x^2-9x)/(2-11x-7x^3) where x approaches infinity.

embizze | High School Teacher | (Level 2) Educator Emeritus

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Evaluate `lim_(x->oo)(6x^3-7x^2-9x)/(2-11x-7x^3)` :

Multiply by `(1/x^3)/(1/x^3)` to get

`=lim_(x->oo)(6-7/x-9/x^2)/(2/x^3-11/x^2-7)`

`=(lim_(x->oo)6-7/x-9/x^2)/(lim_(x->oo)2/x^3-11/x^2-7)`

`=6/(-7)`

So `lim_(x->oo)(6x^3-7x^2-9x)/(2-11x-7x^3)=-6/7`

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This is the horizontal asymptote of the graph of `y=(6x^3-7x^2-9x)/(2-11x-7x^3)`

From Algebra we know that a rational function has a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator.

If the degree of the numerator is less than the degree of the denominator, the asymptote is y=0.

If the degrees are equal, then the asymptote is `y=a/b` where a is the coefficient of the highest degree term in the numerator, and b is the highest degree term in the denominator. The graph:

With calculus you could also apply L'hopital's rule.