# What is the limit of (5x-9)/(4x^3+1) as x goes to infinity?

Oct 9, 2015

0

#### Explanation:

It's clear from the graph shown below that as x gets bigger and bigger, the line gets closer and closer to 0. I know it's not very rigorous, but it works.

graph{(5x-9)/(4x^3+1) [-10, 10, -5, 5]}

Oct 9, 2015

See the explanation section.

#### Explanation:

For all $x$ other than $0$, we have:

$\frac{5 x - 9}{4 {x}^{3} + 1} = \frac{{x}^{3} \left(\frac{5}{x} ^ 2 - \frac{9}{x} ^ 3\right)}{{x}^{3} \left(4 + \frac{1}{x} ^ 3\right)}$

$= \frac{\frac{5}{x} ^ 2 - \frac{9}{x} ^ 3}{4 + \frac{1}{x} ^ 3}$

${\lim}_{x \rightarrow \infty} \left(\frac{5}{x} ^ 2 - \frac{9}{x} ^ 3\right) = 0$

and
${\lim}_{x \rightarrow \infty} \left(4 + \frac{1}{x} ^ 3\right) = 4$.

So,

${\lim}_{x \rightarrow \infty} \frac{5 x - 9}{4 {x}^{3} + 1} = {\lim}_{x \rightarrow \infty} \frac{\frac{5}{x} ^ 2 - \frac{9}{x} ^ 3}{4 + \frac{1}{x} ^ 3}$

$= \frac{0}{4} = 0$