Would you show me how to find the horizontal asymptote for this problem?

Question

f(x)=(3x^3-x^2+x-7)/(2x^3+4x-5)

I know what the answer is but I don't know how to do it. Thanks.

Answer 1

as x gets larger and larger, think of, like a million, the 7 and 5 on the ends don't matter so much. This is pretty much like 3x^3-x^2 +x over 2x^3+4x
which is 3x^2 - x +1 over 2x^2 + 4
as x gets larger, the 1 and 4 don't really matter much, so you have
3x - 1 over 2x
again, and you have 3/2

You can tell it's gonna be 3/2 because the highest power of x is the same on the top and the bottom, and 3 and 2 are the coefficients of the highest power of x.

Answer 2

calculate limit:
lim (3x^3-x^2+x-7)/(2x^3+4x-5)=
x>+-inf

= lim (3-1/x+1/x^2-7/x^3)/(2+4/x^2-
x>+-inf

-5/x^3)=3/2

So horizontal asymptote is y=1.5

Answer 3

y = 3/2, since lim f(x) = 3/2 as x approaches infinity.