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

Question

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!

Answer 1

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 $a=1, n=\frac13$

$$\text{Also observe that }\lim_{h\to0}\frac{(a+h)^n-a^n}h=\frac{d x^n}{dx}_{(\text{at }x=a)}$$

Answer 2

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 $\frac{1}{3}$.

Answer 3

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}

Answer 4

$\displaystyle(x+1)^{1/3}-x^{1/3}=\frac{1}{3c^{2/3}},c\in(x,x+1)$(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 $\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}=1/3=\lim_{x\to \infty}\frac{1}{3}$ )