# How do you find the limit of  (8x^2)/(4x^2-3x-1) as x approaches infinity?

May 29, 2016

2

#### Explanation:

Divide terms on numerator/denominator by ${x}^{2}$

$\frac{\frac{8 {x}^{2}}{x} ^ 2}{\frac{4 {x}^{2}}{x} ^ 2 - \frac{3 x}{x} ^ 2 - \frac{1}{x} ^ 2} = \frac{8}{4 - \frac{3}{x} - \frac{1}{x} ^ 2}$

as $x \to \infty , \frac{3}{x} \text{ and } \frac{1}{x} ^ 2 \to 0$

$\Rightarrow {\lim}_{x \to \infty} \frac{8 {x}^{3}}{4 {x}^{2} - 3 x - 1} = \frac{8}{4} = 2$