# What is the limit of (x^(6) + 4x^(4) - 8x^(2)) / [(x^(4) + 12x^(3) + 2x^(2) + 8)]^5 as x goes to infinity?

Oct 10, 2015

The limit os $0$

#### Explanation:

$\frac{{x}^{6} + 4 {x}^{4} - 8 {x}^{2}}{{x}^{4} + 12 {x}^{3} + 2 {x}^{2} + 8} ^ 5$ has, as its denominator, a degree 20 polynomial with leading coefficient $1$. Because the degree of the denominator is greater than that of the numerator, we have
$\lim \left(x \rightarrow \infty\right) \frac{{x}^{6} + 4 {x}^{4} - 8 {x}^{2}}{{x}^{4} + 12 {x}^{3} + 2 {x}^{2} + 8} ^ 5 = 0$

If we want to write some justification, we can use:

$\frac{{x}^{6} + 4 {x}^{4} - 8 {x}^{2}}{{x}^{4} + 12 {x}^{3} + 2 {x}^{2} + 8} ^ 5 = \frac{{x}^{6} + 4 {x}^{4} - 8 {x}^{2}}{{x}^{4} \left(1 + \frac{12}{x} + \frac{2}{x} ^ 2 + \frac{8}{x} ^ 4\right)} ^ 5$ for $x \ne 0$

$= \frac{{x}^{20} \left(\frac{1}{x} ^ 14 + \frac{4}{x} ^ 16 - \frac{8}{x} ^ 18\right)}{{x}^{20} {\left(1 + \frac{12}{x} + \frac{2}{x} ^ 2 + \frac{8}{x} ^ 4\right)}^{5}}$

$= \frac{\frac{1}{x} ^ 14 + \frac{4}{x} ^ 16 - \frac{8}{x} ^ 18}{1 + \frac{12}{x} + \frac{2}{x} ^ 2 + \frac{8}{x} ^ 4} ^ 5$

Now take limits as $x \rightarrow \infty$, to get

$\frac{0 + 0 + 0}{1 + 0 + 0 + 0} ^ 5 = 0$