# How do you find the limit sqrt(x^2+x)-sqrt(x^2-x) as x->oo?

May 21, 2018

$\textcolor{b l u e}{1}$

#### Explanation:

$\sqrt{{x}^{2} + x} - \sqrt{{x}^{2} - x} \implies \frac{\sqrt{{x}^{2} + x} - \sqrt{{x}^{2} - x}}{1}$

Multiply by the conjugate: $\sqrt{{x}^{2} + x} + \sqrt{{x}^{2} - x}$

$\frac{\left(\sqrt{{x}^{2} + x} + \sqrt{{x}^{2} - x}\right) \left(\sqrt{{x}^{2} + x} - \sqrt{{x}^{2} - x}\right)}{\sqrt{{x}^{2} + x} + \sqrt{{x}^{2} - x}}$

Expand numerator:

$\frac{\left(\sqrt{{x}^{2} + x} + \sqrt{{x}^{2} - x}\right) \left(\sqrt{{x}^{2} + x} - \sqrt{{x}^{2} - x}\right)}{\sqrt{{x}^{2} + x} + \sqrt{{x}^{2} - x}}$

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

$= \frac{2 x}{x \sqrt{1 + \frac{1}{x}} + x \sqrt{1 - \frac{1}{x}}}$

Cancelling:

$= \frac{2}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}}$

as $x \to \infty \textcolor{w h i t e}{8888}$, $\frac{2}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}} \to \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}}$

$= \frac{2}{2} = 1$

$\therefore$

${\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + x} - \sqrt{{x}^{2} - x}\right) = 1$