# How do you find the limit of (8x^3 + 5x^2)^(1/3) -2x  as x approaches infinity?

Oct 21, 2016

Rationalize and simplify.

#### Explanation:

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

This allows us to rationalize $\sqrt[3]{u} - v$ by multiplying by

${\left(\sqrt[3]{u}\right)}^{2} + v \sqrt[3]{u} + {v}^{2}$ to get $u - {v}^{3}$.

We'll do that with $u = \sqrt[3]{8 {x}^{3} + 5 {x}^{2}}$ and $v = 2 x$.

$\frac{\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}} - 2 x\right)}{1} \cdot \frac{\left({\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}}\right)}^{2} + 2 x \sqrt[3]{8 {x}^{3} + 5 {x}^{2}} + {\left(2 x\right)}^{2}\right)}{\left({\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}}\right)}^{2} + 2 x \sqrt[3]{8 {x}^{3} + 5 {x}^{2}} + {\left(2 x\right)}^{2}\right)}$

$= \frac{{\left(\sqrt[3]{8 {x}^{2} + 5 {x}^{2}}\right)}^{3} - {\left(2 x\right)}^{3}}{\left({\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}}\right)}^{2} + 2 x \sqrt[3]{8 {x}^{3} + 5 {x}^{2}} + {\left(2 x\right)}^{2}\right)}$

$= \frac{8 {x}^{2} + 5 {x}^{2} - 8 {x}^{3}}{\left({\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}}\right)}^{2} + 2 x \sqrt[3]{8 {x}^{3} + 5 {x}^{2}} + 4 {x}^{2}\right)}$

$= \frac{5 {x}^{2}}{\left({\left(\sqrt[3]{8 {x}^{3} + 5 {x}^{2}}\right)}^{2} + 2 x \sqrt[3]{8 {x}^{3} + 5 {x}^{2}} + 4 {x}^{2}\right)}$

Now we'll do some work in the denominator to factor out an ${x}^{2}$.

For $x \ne 0$, we have

 = (5x^2)/ (((root(3)(x^3(8 + 5/x)))^2+2x root(3)(x^3(8 + 5/x) + 4x^2))

$= \frac{5 {x}^{2}}{\left({\left(x \sqrt[3]{\left(8 + \frac{5}{x}\right)}\right)}^{2} + 2 {x}^{2} \sqrt[3]{8 + \frac{5}{x}} + 4 {x}^{2}\right)}$

$= \frac{5 {x}^{2}}{{x}^{2} \left({\left(\sqrt[3]{\left(8 + \frac{5}{x}\right)}\right)}^{2} + 2 \sqrt[3]{8 + \frac{5}{x}} + 4\right)}$

$= \frac{5}{{\left(\sqrt[3]{\left(8 + \frac{5}{x}\right)}\right)}^{2} + 2 \sqrt[3]{8 + \frac{5}{x}} + 4}$.

Evaluating the limit as $x \rightarrow \infty$, we finish with

$= \frac{5}{{\left(\sqrt[3]{\left(8 + 0\right)}\right)}^{2} + 2 \sqrt[3]{8 + 0} + 4}$

$= \frac{5}{4 + 4 + 4} = \frac{5}{12}$.