Question

Find the limit of rational function as x->infinity and as x->-infinity
h(x)=(-2x^3-2x+3)/(3x^3+3x^2-5x)

Answer 1

Divide everything by \(x^3\) (in general, the highest power of \(x\))
When you have all your \(x\)s int he denominator, the limit will work out just fine.

Answer 2

@wio do i put x^3 at the bottom?

Answer 3

\[\large
h(x)=\frac{-2x^3-2x+3}{3x^3+3x^2-5x} = \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{5}{x^2}}
\]

Answer 4

When you set \(x\to \infty\) you'll notice that most of the terms go to \(0\)

Answer 5

I am confused how did you set this up I mean the top and bottom?

Answer 6

Remember that \[\large
1 = \frac{x^{-3}}{x^{-3}} = \frac{\frac{1}{x^3}}{\frac{1}{x^3}}
\]I just multiplied it by \(1\).