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

May 5, 2016

The limit is $\frac{1}{2}$. For "how", please see the explanation section below.

#### Explanation:

In investigating a limit as $x$ increases without bounds, we are not concerned about what happens at $x = 0$.

We note that for all $x \ne 0$,

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

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

Note also that $\sqrt{{x}^{2}} = \left\mid x \right\mid$, so for positive $x$, we have $\sqrt{{x}^{2}} = x$.

(For negative $x$, we get $\sqrt{{x}^{2}} = - x$)

We continue:

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

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

Evaluating the limit as $x \rightarrow \infty$, we get

${\lim}_{x \rightarrow \infty} \frac{\sqrt{{x}^{2} - 1}}{\sqrt{4 {x}^{2} + x}} = \frac{\sqrt{1 - 0}}{\sqrt{4 + 0}} = \frac{1}{2}$