SOLUTION: Give the equation of the horizontal asymptote, if any, of the function. G(x) = x(x-1)/x^3+49x A) y = 0 B) y = 1 C) x = 0, x = -49 D) None I worked it out like

Question 437663: Give the equation of the horizontal asymptote, if any, of the function.
G(x) = x(x-1)/x^3+49x
A) y = 0 B) y = 1 C) x = 0, x = -49 D) None
I worked it out like this: x(x-1)/x^3+49x = x^2-x/x(x^2+49) x^2/x^2 y = 1
However, I am not sure if that is correct. What are the correct steps to solve this and what is the right answer?
Thank you very much for your help.
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
The degree of the denominator is greater than the degree of the numerator, and so for x-values close to infinity, the two dominant terms are related
as , which would approach 0 as x goes to infinity. The horizontal asymptote is thus y = 0.
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