How do you find the limit of x^5/(4x^7-x^3+9) as x approaches infinity?

Aug 4, 2017

The limit as $x \to \infty$ is $= 0$

Explanation:

There are $2$ ways for calculating this limit

First method

We take the term of highest degree in the numerator and in the denominator

${\lim}_{x \to \pm \infty} \frac{{x}^{5}}{4 {x}^{7} - {x}^{3} + 9} = {\lim}_{x \to \pm \infty} \frac{{x}^{5}}{4 {x}^{7}}$

$= {\lim}_{x \to \pm \infty} \frac{1}{4 {x}^{2}} = 0$

Second method

Divide the denominator by ${x}^{7}$

$\frac{{x}^{5}}{4 {x}^{7} - {x}^{3} + 9} = \frac{{x}^{5}}{{x}^{7} \left(4 - \frac{1}{x} ^ 4 + \frac{9}{x} ^ 7\right)}$

$= \frac{1}{{x}^{2} \left(4 - \frac{1}{x} ^ 4 + \frac{9}{x} ^ 7\right)}$

${\lim}_{x \to \pm \infty} \frac{1}{x} ^ 4 = 0$

${\lim}_{x \to \pm \infty} \frac{9}{x} ^ 7 = 0$

Therefore,

${\lim}_{x \to \pm \infty} \frac{{x}^{5}}{4 {x}^{7} - {x}^{3} + 9} = {\lim}_{x \to \pm \infty} \frac{1}{4 {x}^{2}} = 0$