# What is the limit of root3[x^3+2] - root3[x^3-1] as x goes to infinity?

Sep 26, 2015

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{3} + 2} - \sqrt{{x}^{3} - 1}\right) = 0$

#### Explanation:

Use the difference of cubes to rewrite the expression.

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

So,

$\left(\sqrt{a} - \sqrt{b}\right) \left({\sqrt{a}}^{2} + \sqrt{a b} + {\sqrt{b}}^{2}\right) = a - b$

And, so

$\frac{\sqrt{a} - \sqrt{b}}{1} \cdot \frac{{\sqrt{a}}^{2} + \sqrt{a b} + {\sqrt{b}}^{2}}{{\sqrt{a}}^{2} + \sqrt{a b} + {\sqrt{b}}^{2}} = \frac{a - b}{{\sqrt{a}}^{2} + \sqrt{a b} + {\sqrt{b}}^{2}}$

In this case, we have:

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

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

As $x$ increases without bound, the denominator also increases without bound, so the limit of the ratio is $0$.