# How do you find the limit of (2 - root3x) / (sqrt(x - 4) - 2) as x approaches 8?

Mar 26, 2016

Use the conjugates of the numerator and the denominator to find that the limit is $- \frac{1}{3}$.

#### Explanation:

The conjugate of $\sqrt{u} - v$ is $\sqrt{u} + v$ because the product, $u - {v}^{2}$ has no radicals. (That is not the definition of conjugate, but it works for this question.)

We can understand why this works by recalling from more basic algebra, that $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$. So if one, or both, of $a$, $b$ involve a square (second) root, then their squares do not involve roots.

Consider, then, third powers and third roots.

I hope that you learned in your previous study of algebra that

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ and ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

Here, if $a$ or $b$ or both involve cube (third) roots, then the cubes do not involve roots.

On to the question at hand:

${\lim}_{x \rightarrow 8} \frac{2 - \sqrt[3]{x}}{\sqrt{x - 4} - 2}$

If we try to evaluate by substitution, we get the indeterminate form $\frac{0}{0}$.
We'll use the conjugates to rewrite the quotient:

$\frac{2 - \sqrt[3]{x}}{\sqrt{x - 4} - 2} \cdot \frac{\left({2}^{2} + 2 \sqrt[3]{x} + \sqrt[3]{{x}^{2}}\right)}{\left({2}^{2} + 2 \sqrt[3]{x} + \sqrt[3]{{x}^{2}}\right)} \cdot \frac{\left(\sqrt{x - 4} + 2\right)}{\left(\sqrt{x - 4} + 2\right)}$

$= \frac{\left({2}^{3} - x\right) \left(\sqrt{x - 4} + 2\right)}{\left(\left(x - 4\right) - 4\right) \left({2}^{2} + 2 \sqrt[3]{x} + \sqrt[3]{{x}^{2}}\right)}$

$= \frac{\left(8 - x\right) \left(\sqrt{x - 4} + 2\right)}{\left(x - 8\right) \left({2}^{2} + 2 \sqrt[3]{x} + \sqrt[3]{{x}^{2}}\right)}$

$= \frac{- 1 \left(\sqrt{x - 4} + 2\right)}{\left({2}^{2} + 2 \sqrt[3]{x} + \sqrt[3]{{x}^{2}}\right)}$.

We can now evaluate the limit by substitution. The form is no longer indeterminate.

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

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