# How do you find the limit of sqrt(1- 8x^3) /sqrt(1-4x^2) as x approaches 1/2?

Apr 24, 2016

${\lim}_{x \rightarrow \frac{1}{2} {\textcolor{w h i t e}{}}^{-}} \frac{\sqrt{1 - 8 {x}^{3}}}{\sqrt{1 - 4 {x}^{2}}} = \frac{\sqrt{6}}{2}$

#### Explanation:

We will use the following identities, which are a difference of cubes and a difference of squares.

• $1 - 8 {x}^{3} = {1}^{3} - {\left(2 x\right)}^{3} = \left(1 - 2 x\right) \left(1 + 2 x + 4 {x}^{2}\right)$
• $1 - 4 {x}^{2} = 1 - {\left(2 x\right)}^{2} = \left(1 + 2 x\right) \left(1 - 2 x\right)$

Thus, we have the limit

${\lim}_{x \rightarrow \frac{1}{2}} \frac{\sqrt{1 - 8 {x}^{3}}}{\sqrt{1 - 4 {x}^{2}}} = {\lim}_{x \rightarrow \frac{1}{2}} \frac{\sqrt{\left(1 - 2 x\right) \left(1 + 2 x + 4 {x}^{2}\right)}}{\sqrt{\left(1 + 2 x\right) \left(1 - 2 x\right)}}$

Square roots can be split up through multiplication:

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

Cancel the $\sqrt{1 - 2 x}$ terms:

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

We can now evaluate the limit by plugging in $\frac{1}{2}$ for $x$:

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

This can also be written as $\frac{\sqrt{6}}{2}$.

Note that $\frac{\sqrt{6}}{2} \approx 1.225$ and examine the graph of the function:

graph{sqrt(1-8x^3)/sqrt(1-4x^2) [-2.827, 2.648, -0.39, 2.347]}

The graph of the function appears to be approaching $\approx 1.225$ at $x = \frac{1}{2}$.

Note:

Upon examining the graph, it becomes clear that this is a one-sided limit (from the left). This occurs since for $x > \frac{1}{2}$, we have a negative value inside the square root.

Thus, the accurate statement to make is that the limit from the left approaches $\frac{\sqrt{6}}{2}$, expressed mathematically as:

${\lim}_{x \rightarrow \frac{1}{2} {\textcolor{w h i t e}{}}^{-}} \frac{\sqrt{1 - 8 {x}^{3}}}{\sqrt{1 - 4 {x}^{2}}} = \frac{\sqrt{6}}{2}$