# What is the limit as x approaches infinity of (x^2-4)/(2+x-4x^2)?

Sep 28, 2014

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

Let us look at some details.
There are a couple of methods to do this.

Method 1

${\lim}_{x \to \infty} \frac{{x}^{2} - 4}{2 + x - 4 {x}^{2}}$

by dividing the numerator and the denominator by ${x}^{2}$,
(${x}^{2}$ was chosen to match the leading term in the denominator.)

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

Method 2

${\lim}_{x \to \infty} \frac{{x}^{2} - 4}{2 + x - 4 {x}^{2}}$

by l'Hopital's Rule ($\infty$/$\infty$),

$= {\lim}_{x \to \infty} \frac{2 x}{1 - 8 x}$

by l'Hopital's Rule ($\infty$/$\infty$) again,

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

I hope that this was helpful.