# Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$

Need to solve following problem: $$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$$ I've tried to do something like this: $$\lim_{x \to \infty} x\left(\sqrt[3]{5+8x^3} - 2x\right) =\lim_{x \to \infty}x\left( \sqrt[3]{ \left( \frac {5}{x^3}+8 \right)x^3 } - 2x\right) = \lim_{x \to \infty} x\left( \left ( \sqrt[3] { \frac{5}{x^3}+8} \right)x - 2x\right)$$ It seems to be the right way, but I can't do my next step.

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

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Note that $$2x\lt \sqrt[3]{5+8x^3}=2x\left(1+\frac{5}{8x^3}\right)^{1/3}\lt 2x\left(1+\frac{5}{24x^3}\right).$$ Then use Squeezing.

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The Squeezing is nice but I can't use it now, because I am first-year student. – k1ber Oct 13 '13 at 20:48
The identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ used by njguliev works nicely. – André Nicolas Oct 13 '13 at 20:52
But I don't understand what I have to do next. Could you explain me please? – k1ber Oct 13 '13 at 20:56
You mean in the answer by njguliev? Multiply by $x$. The top becomes $5x$, and the bottom is clearly bigger than $(2x)^2$, indeed bigger than $4(2x)^2$, so the limit is $0$. – André Nicolas Oct 13 '13 at 21:07

Putting $n=\frac1h$

$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$$

$$=\lim_{h\to0}\frac{\sqrt[3]{5h^3+8}-2}{h^2}$$

$$=\lim_{h\to0}\frac{5h^3+8-2^3}{h^2\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$

$$=\lim_{h\to0}\frac{5h}{\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$

$$=\cdots$$

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