# Find Limit of sqrt (8x^2 + 4x - 8) - sqrt (2x^2 - 3x+1) - sqrt (2x^2 + 4x-3) as x->infinity?

## how to rationalize

Mar 8, 2018

$\frac{1}{2 \sqrt{2}}$

#### Explanation:

$\sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} - 3 x + 1} - \sqrt{2 {x}^{2} + 4 x - 3} =$

$\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} - 3 x + 1} + \frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} + 4 x - 3}$

Now

$\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} - 3 x + 1} = \left(\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} - 3 x + 1}\right) \frac{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} - 3 x + 1}}{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} - 3 x + 1}} =$

$= \frac{4 x - 3}{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} - 3 x + 1}}$

and

$\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} + 4 x - 3} = \left(\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} + 4 x - 3}\right) \frac{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} + 4 x - 3}}{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} + 4 x - 3}} =$

$\frac{1 - 3 x}{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} + 4 x - 3}}$

Now

${\lim}_{x \to \infty} \frac{4 x - 3}{\frac{1}{2} \sqrt{8 {x}^{2} + 4 x - 8} + \sqrt{2 {x}^{2} - 3 x + 1}} =$

${\lim}_{x \to \infty} \frac{4 - \frac{3}{x}}{\frac{1}{2} \sqrt{8 + \frac{4}{x} - \frac{8}{x} ^ 2} + \sqrt{2 - \frac{3}{x} + \frac{1}{x} ^ 2}} = \frac{4}{2 \sqrt{2}}$

analogously we can proceed obtaining the final result

${\lim}_{x \to \infty} \sqrt{8 {x}^{2} + 4 x - 8} - \sqrt{2 {x}^{2} - 3 x + 1} - \sqrt{2 {x}^{2} + 4 x - 3} = \frac{1}{2 \sqrt{2}}$