# How do you find the limit of (x^3 + 1) / (sqrt { 4x^2 + 3} ) as x approaches infinity?

Jun 20, 2015

Use $\sqrt{4 {x}^{2} + 3} = \sqrt{{x}^{2}} \sqrt{4 + \frac{3}{x} ^ 2}$, for $x \ne 0$ and $= x \sqrt{4 + \frac{3}{x} ^ 2}$ for $x > 0$

#### Explanation:

This answer assumes that you either do not have l'Hopital's Rule, or you prefer not to use it.

Note: $\sqrt{{x}^{2}} = \left\mid x \right\mid = \left\{\begin{matrix}x & \text{ if " & x >=0 \\ -x & " if } & x < 0\end{matrix}\right.$

To find ${\lim}_{x \rightarrow \infty}$, we are only interested in positive values of $x$, so we can use:

(x^3 + 1) / (sqrt { 4x^2 + 3} ) = (x^3+1)/sqrt(x^2(4+3/x^2)

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

$= \frac{x \left({x}^{2} + \frac{1}{x}\right)}{x \sqrt{4 + \frac{3}{x} ^ 2}}$ (for $x > 0$)

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

Clearly, this last ratio approaches ${x}^{2} / 2$ as $x \rightarrow \infty$, so the limit is $\infty$

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