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!
 A: Putting $h=\frac1x$ as $x\to\infty, h\to 0$
$$\lim_{x\to\infty}x^\frac23 \left\{(x+1)^\frac13-x^\frac13 \right\}$$
$$\lim_{h\to0}\frac1{h^\frac23} \left\{\left(1+\frac1h\right)^\frac13-\frac1{h^\frac13} \right\}$$
$$=\lim_{h\to0}\frac{(1+h)^\frac13-1}h$$
$$=\lim_{h\to0}\frac{1+\frac13h +O(h^2)-1}h$$ using the Generalized Binomial Theorem 
$$=\lim_{h\to0} \left\{\frac13+O(h) \right\}$$ 
$$=\frac13$$
A little generalization :
$$\lim_{h\to0}\frac{(a+h)^n-a^n}h\text{ where } n \text{ finite non-zero real number}$$
$$=a^n\lim_{h\to0}\frac{(1+\frac ha)^n-1}h$$ 
$$=a^n\lim_{h\to0}\frac{(1+n\frac ah+O(h^2))-1}h$$ as $h\to0,$ we can always set $|\frac ha|<1$ to utilize the Generalized Binomial Theorem
$$=na^{n-1}+\lim_{h\to0} O(h)\text { as } h\ne 0\text{ as }h\to0$$
$$=na^{n-1}$$
Putting $x=\frac1h$   as $h\to0,x\to\infty$
$$\lim_{h\to0}\frac{(a+h)^n-a^n}h=\lim_{x\to\infty}\frac{(a+\frac1x)^n-a^n}{\frac1x}\lim_{x\to\infty}x^{1-n}\{(ax+1)^n-(ax)^n\}$$
Here $a=1, n=\frac13$
$$\text{Also observe that }\lim_{h\to0}\frac{(a+h)^n-a^n}h=\frac{d x^n}{dx}_{(\text{at }x=a)}$$
A: Now the limit makes sense. I think the best way to see it is to expand $(x + 1)^{1/3}$ by using either a generalized binomial theorem or Taylor expansion (whichever you feel more comfortable with).
In each case, we see that $(x + 1)^{1/3} = x^{1/3} + \frac{1}{3}x^{-2/3} + \frac{1}{3}\frac{-2}{3}x^{-5/3} + \text{lower order terms}$.
Thus $$\begin{align}x^{2/3}((x+1)^{1/3} - x^{1/3}) &= x^{2/3}(-x^{1/3} + x^{1/3} + \frac{1}{3}x^{-2/3} + \text{lower order terms})\\ &= \frac{1}{3} + \text{lower order terms}\end{align}$$
where I use 'lower order terms' to indicate things that all go to zero when $x \to \infty$, even when multiplied by $x^{2/3}$ as is this case.
Thus the limit is $\frac{1}{3}$.
A: Probably you meant
$$\lim _{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
Which can dealt using the identity
$$x^3-y^3=(x-y)(x^2+y^2+xy)$$
now apply it for $x=x^{\frac{1}{3}}, \,y=(x+1)^{\frac{1}{3}}$
so multiply the numerator and denominator with
$x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}$
and you will get:
\begin{align}\lim _{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})&=\lim _{x\to\infty}x^{\frac{2}{3}}\frac{(x+1)^{1/3}-x^{1/3})(x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}) }{x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}}\\
&=\lim _{x\to\infty}x^\frac{2}{3}\frac{1}{x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}},\mathrm{now\,divide\,with \, x^\frac{2}{3}}\\
&=\lim _{x\to\infty}\frac{1}{1+(1+\frac{1}{x})^\frac{2}{3}+\left (\frac{x^2+x}{x^2}\right )^\frac{1}{3}}\\
&=\frac{1}{3}
\end{align}
A: $\displaystyle(x+1)^{1/3}-x^{1/3}=\frac{1}{3c^{2/3}},c\in(x,x+1)$(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 $\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}=1/3=\lim_{x\to \infty}\frac{1}{3}$ )
