# What is the limit of sqrt(4x^2-1) / x^2 as x goes to negative infinity?

Sep 26, 2015

The limit is zero.

#### Explanation:

As $x$ approaches $- \setminus \infty$, the $- 1$ inside the root becomes insignificant (which is to say, $4 {x}^{2}$ and $4 {x}^{2} - 1$ are asymptotically equivalent, which again means that you can substitute one with the other and the limits doesn't change).

So, you have the following:

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

With the advantage that $\setminus \sqrt{4 {x}^{2}}$ is easibly computable, since it is $\left\mid 2 x \right\mid$. And since $\frac{\left\mid 2 x \right\mid}{x} ^ 2 = \frac{2}{\left\mid x \right\mid}$, you can clearly see that this function goes to $0$, as $x$ approaches $- \setminus \infty$.

Sep 26, 2015

Here is a different way of writing the solution.

#### Explanation:

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

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

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

As $x \rightarrow - \infty$ we are concerned with negative values of $x$,.

For $x < 0$, $\sqrt{{x}^{2}} = \left\mid x \right\mid = - x$.

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

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

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

$= 0$.

(I would describe this as more algebraic and less analytic than the other fine answer provided by Killer Bunny (and George C.).)