# Math Help - Determing a limit

1. ## Determing a limit

The question states: Find the limit of the sequence
{ $n^{2/3}[(n+1)^{1/3}-n^{1/3}]$}

I'm assuming they are asking as n approaches infinity, would anyone be able to offer tips on where to begin to solve?

2. $\lim_{n\to\infty}\sqrt[3]{n^2}\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)=$

$=\lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n^2+n}+\sqrt[3]{n^2}}=$

$\lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{n^2}\left(\sqrt[3]{\left(1+\frac{1}{n}\right)^2}+\sqrt[3]{1+\frac{1}{n}}+1\right)}=$

$=\lim_{n\to\infty}\frac{1}{\sqrt[3]{\left(1+\frac{1}{n}\right)^2}+\sqrt[3]{1+\frac{1}{n}}+1}=\frac{1}{3}$

3. red dog used the fact that $a^3- b^3= (a- b)(a^2+ ab+ b^2)$ with $a= (n+1)^{1/3}$ and $b= n^{1/3}$.

4. anoher approach could be by putting $t=\frac1n$ and then put $t+1=u^3.$

5. Thanks guys!
one step I don't seem to fully understand is going from:
$\lim_{n\to\infty}\sqrt[3]{n^2}\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)$
to
$\lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n^2+n}+\sqrt[3]{n^2}}$
why does part of the factored expression all move into the denominator?

6. Originally Posted by xxlvh
The question states: Find the limit of the sequence
{ $n^{2/3}[(n+1)^{1/3}-n^{1/3}]$}

I'm assuming they are asking as n approaches infinity, would anyone be able to offer tips on where to begin to solve?
$n^{2/3}[(n+1)^{1/3}-n^{1/3}]=n[(1+1/n)^{1/3}-1]=n\left[\left(1+\frac{1}{3n}+O(n^{-2})\right) -1\right]$ $=1/3+O(n^{-1})$

7. $\underset{n\to \infty }{\mathop{\lim }}\,\left( \sqrt[3]{n^{3}+n^{2}}-n \right),$ as i said before, put $t=\frac1n$ and the limit becomes $\underset{t\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{t+1}-1}{t},$ and then put $u=\sqrt[3]{t+1}$ and the limit is $\underset{u\to 1}{\mathop{\lim }}\,\frac{u-1}{u^{3}-1}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1}{u^{2}+u+1}=\frac{1}{3}.$