# How do you find the limit lim (2x^2-3x+2)/(x^3+5x^2-1) as x->oo?

Dec 15, 2016

${\lim}_{x \rightarrow \infty} \frac{2 {x}^{2} - 3 x + 2}{{x}^{3} + 5 {x}^{2} - 1} = 0$

#### Explanation:

Note that as $x \rightarrow \infty$ then $\frac{1}{x} , \frac{1}{x} ^ 2 , \frac{1}{x} ^ 3 \rightarrow 0$

We can therefore multiply numerator and denominator by $\frac{1}{x} ^ 3$ (the reciprocal of the largest power in the denominator) as follows:

${\lim}_{x \rightarrow \infty} \frac{2 {x}^{2} - 3 x + 2}{{x}^{3} + 5 {x}^{2} - 1} = {\lim}_{x \rightarrow \infty} \frac{2 {x}^{2} - 3 x + 2}{{x}^{3} + 5 {x}^{2} - 1} \cdot \frac{\frac{1}{x} ^ 3}{\frac{1}{x} ^ 3}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{\left(\frac{1}{x} ^ 3\right) \left(2 {x}^{2} - 3 x + 2\right)}{\left(\frac{1}{x} ^ 3\right) \left({x}^{3} + 5 {x}^{2} - 1\right)}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{\frac{2}{x} - \frac{3}{x} ^ 2 + \frac{2}{x} ^ 3}{1 + \frac{5}{x} - \frac{1}{x} ^ 3}$
$\text{ } = \frac{0 - 0 + 0}{1 + 0 - 0}$
$\text{ } = 0$

We can verify this result by looking at the graph of $y = \frac{2 {x}^{2} - 3 x + 2}{{x}^{3} + 5 {x}^{2} - 1}$
graph{(2x^2-3x+2)/(x^3+5x^2-1) [-7, 13, -4.16, 5.84]}
and indeed it does appear that for large $x$ the function is approaching a horizontal asymptote $y = 0$