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

Apr 15, 2016

Use $\left(\sqrt{a} - b\right) \left(\sqrt{a} + b\right) = a - {b}^{2}$ and $\sqrt{{u}^{2}} = \left\mid u \right\mid$ and some other algebra.

#### Explanation:

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

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

$= \frac{x}{\sqrt{{x}^{2} \left(1 + \frac{1}{x}\right)} + x}$ $\text{ }$ (for $x \ne 0$)

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

When evaluating the limit as $x$ increases without bound, we are concerned only with positive values of $x$.

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

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

 = x/(x(sqrt(1+1/x) +1)

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

So, finally, we get:

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

$= \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$