<s>[INST] Prove: $\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ How can I show: 

$$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$

I've tried multiplying with its "conjugate" but that doesn't seem to help that much.
Thanks!
 [/INST] $\displaystyle(x+1)^{1/3}-x^{1/3}=\frac{1}{3c^{2/3}},c\in(x,x+1)$(By mean value theorem)
$\displaystyle \frac{1}{3(x+1)^{2/3}}\le\frac{1}{3c^{2/3}}\le\frac{1}{3x^{2/3}},\forall c\in(x,x+1)$
So we have ,
$\displaystyle\frac{1}{3(x+1)^{2/3}}\le(x+1)^{1/3}-x^{1/3}\le \frac{1}{3x^{2/3}}$
$\Rightarrow \displaystyle\frac{x^{2/3}}{3(x+1)^{2/3}}\le x^{2/3}((x+1)^{1/3}-x^{1/3})\le \frac{1}{3}$
$\Rightarrow \displaystyle\frac{1}{3(1+\frac{1}{x})^{2/3}}\le x^{2/3}((x+1)^{1/3}-x^{1/3})\le \frac{1}{3}$
Taking limit as $x\to \infty$ we have,
$\Rightarrow \displaystyle\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}\le \lim_{x\to \infty}x^{2/3}((x+1)^{1/3}-x^{1/3})\le \lim_{x\to \infty}\frac{1}{3}$
Using squeeze theorem, $\Rightarrow \lim_{x\to \infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ (As $\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}=1/3=\lim_{x\to \infty}\frac{1}{3}$ )
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