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

Sep 9, 2016

Please see the explanation section below.

#### Explanation:

$\sqrt{4 {x}^{2} + 3 x} - 2 x = \frac{\left(\sqrt{4 {x}^{2} + 3 x} - 2 x\right)}{1} \cdot \frac{\left(\sqrt{4 {x}^{2} + 3 x} + 2 x\right)}{\left(\sqrt{4 {x}^{2} + 3 x} + 2 x\right)}$

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

$= \frac{3 x}{\sqrt{{x}^{2}} \sqrt{4 + \frac{3}{x}} + 2 x}$ $\text{ }$ for $x \ne 0$

For $x > 0$, we have $\sqrt{{x}^{2}} = x$, so

$= \frac{3 x}{x \sqrt{4 + \frac{3}{x}} + 2 x}$ $\text{ }$ for $x > 0$

$= \frac{3 x}{x \left(\sqrt{4 + \frac{3}{x}} + 2\right)}$ $\text{ }$ for $x > 0$

$= \frac{3}{\sqrt{4 + \frac{3}{x}} + 2}$ $\text{ }$ for $x > 0$.

Now as $x \rightarrow \infty$, we get

$\frac{3}{\sqrt{4 + 0} + 2} = \frac{3}{4}$