# How do you find the limit of sqrt(9x^2 +x)-(3x) as x approaches infinity?

Sep 6, 2016

$= \frac{1}{6}$

#### Explanation:

${\lim}_{x \to \infty} \sqrt{9 {x}^{2} + x} - 3 x$

${\lim}_{x \to \infty} 3 x \sqrt{1 + \frac{1}{9 x}} - 3 x$

${\lim}_{x \to \infty} 3 x \left({\left(1 + \frac{1}{9 x}\right)}^{\frac{1}{2}} - 1\right)$

by Binomial Expansion

${\lim}_{x \to \infty} 3 x \left(1 + \frac{1}{2} \frac{1}{9 x} + O \left(\frac{1}{x} ^ 2\right) - 1\right)$

${\lim}_{x \to \infty} 3 x \left(\frac{1}{18 x} + O \left(\frac{1}{x} ^ 2\right)\right)$

${\lim}_{x \to \infty} \frac{1}{6} + O \left(\frac{1}{x}\right) = \frac{1}{6}$

Sep 6, 2016

$\frac{1}{6}$

#### Explanation:

$\frac{\left(\sqrt{9 {x}^{2} + x} - 3 x\right)}{1} \cdot \frac{\left(\sqrt{9 {x}^{2} + x} - 3 x\right)}{\left(\sqrt{9 {x}^{2} + x} - 3 x\right)} = \frac{9 {x}^{2} + x - 9 {x}^{2}}{\left(\sqrt{9 {x}^{2} + x} - 3 x\right)}$

$= \frac{x}{\sqrt{9 {x}^{2} + x} - 3 x}$

$= \frac{x}{\sqrt{{x}^{2}} \sqrt{9 + \frac{1}{x}} - 3 x}$ $\text{ }$ for $x \ne 0$

$= \frac{x}{x \left(\sqrt{9 + \frac{1}{x}} - 3\right)}$

$= \frac{1}{\sqrt{9 + \frac{1}{x}} - 3}$

Now, as $x \rightarrow \infty$, $\frac{1}{x} \rightarrow 0$, so the limit is

$\frac{1}{\sqrt{9} + 3} = \frac{1}{6}$