# How do you find lim (10x^2+x+2)/(x^3-4x^2-1) as x->oo?

Jan 9, 2017

${\lim}_{x \to \infty} \frac{10 {x}^{2} + x + 2}{{x}^{3} - 4 {x}^{2} - 1} = 0$

#### Explanation:

When calculating the limit of a rational function for $x \to \pm \infty$ you can ignore all the monomials above and below the line except the ones with the highest order:

${\lim}_{x \to \infty} \frac{10 {x}^{2} + x + 2}{{x}^{3} - 4 {x}^{2} - 1} = {\lim}_{x \to \infty} \frac{10 {x}^{2}}{x} ^ 3 = {\lim}_{x \to \infty} \frac{10}{x} = 0$

If you want to see why this is the case, split the rational function as a sum and then bring each monomial below the line:

$\frac{10 {x}^{2} + x + 2}{{x}^{3} - 4 {x}^{2} - 1} = \frac{10 {x}^{2}}{{x}^{3} - 4 {x}^{2} - 1} + \frac{x}{{x}^{3} - 4 {x}^{2} - 1} + \frac{2}{{x}^{3} - 4 {x}^{2} - 1} = \frac{10}{\frac{{x}^{3} - 4 {x}^{2} - 1}{x} ^ 2} + \frac{1}{\frac{{x}^{3} - 4 {x}^{2} - 1}{x}} + \frac{2}{{x}^{3} - 4 {x}^{2} - 1} = \frac{10}{x - 4 - \frac{1}{x} ^ 2} + \frac{1}{{x}^{2} - 4 x - \frac{1}{x}} + \frac{2}{{x}^{3} - 4 {x}^{2} - 1}$

Evidently all three addenda are infinitesimal for $x \to \infty$