Evaluate $\lim\limits_{x\to+\infty} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )$? 
How do you evaluate $$\lim\limits_{x\to+\infty} \sqrt{x}\left (\sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) ?$$

Thanks in advance for your support.
 A: $$...=\lim\limits_{x\to\infty}  {2\sqrt{x}\over \sqrt[3]{(x+1)^2}+\sqrt[3]{x^2-1}+\sqrt[3]{(x-1)^2}}$$ $$=\lim\limits_{x\to\infty}  {2\sqrt{x}\over 3\sqrt[3]{x^2}}$$ $$={2\over 3}\lim\limits_{x\to\infty}  {1\over \sqrt[6]{x}}$$ $$=0$$
A: Hint: multiply the numerator and the denominator for $(x+1)^{2/3} + (x-1)^{2/3} + (x^2 -1)^{1/3}$. What happens? 
A: Let $1/x=h^6$ as $[2,3]=6$ 
$$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h^6}-\sqrt[3]{1-h^6}}{h^{3+2}}$$
$$=\lim_{h\to0^+}\dfrac{1+h^6-(1-h^6)}{h^5}\cdot\lim_{h\to0^+}\dfrac1{(1+h^6)^{2/3}+(1+h^6)^{1/3}(1-h^6)^{1/3}+(1-h^6)^{2/3}}$$
$$=0\cdot\dfrac1{(1+1\cdot1+1)}$$
A: If it is ok for you to use only the fact that $\color{blue}{f(t)} =\sqrt[3]{1+t}$ and $\color{blue}{g(t)}=\sqrt[3]{1-t}$ are differentiable at $t= 0$, then without calculating any derivative you may proceed as follows:


*

*Set $x = \frac{1}{t}$ and consider $t \rightarrow 0^+$
\begin{eqnarray*} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )
& \stackrel{x=\frac{1}{t}}{=} & \frac{\sqrt[3]{1+t} - \sqrt[3]{1-t}}{\underbrace{\sqrt[3]{t}\cdot \sqrt{t}}_{=t^{\frac{5}{6}}}}\\
& = & \frac{\sqrt[3]{1+t} -1 + 1 - \sqrt[3]{1-t}}{t}\cdot t^{\frac{1}{6}}\\
& = & \left(\underbrace{\frac{\sqrt[3]{1+t} -1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{f'(0)}} - \underbrace{\frac{\sqrt[3]{1-t} - 1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{g'(0)}}\right) \cdot t^{\frac{1}{6}}\\
& \stackrel{t\to 0^+}{\longrightarrow} & (f'(0) - g'(0))\cdot 0 = 0
\end{eqnarray*}
