Finding the limit of a sequence $\lim _{n\to \infty} \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right)$

Question

If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it.

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

Answer 1

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

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

Answer 2

$$ \begin{align*} \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right) &= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } \frac{\sqrt [3]{1+ \frac{1}{n}}- 1 }{\frac{1}{n}} \\ &= \lim _{h \to 0} \frac{\sqrt [3]{1+ h}- 1 }{h} \\ &= \left. \frac{d}{du} \sqrt[3]{u} \ \right|_{u=1} \\ &= \cdots \end{align*} $$

Answer 3

Presumably you don't want a Taylor series expansion, since you said you "don't want to differentiate anything," but it's worth pointing out that you can apply the binomial expansion: $$ \begin{eqnarray} \sqrt[3]{n+1} &=& \sqrt[3]{n}\sqrt[3]{1+n^{-1}} \\ &=& \sqrt[3]{n}\sum_{k}{{1/3}\choose{k}}n^{-k} \\ &=& \sum_{k}{{1/3}\choose{k}}n^{1/3-k} \\ &=& \sqrt[3]{n} + \frac{1}{3}n^{-2/3}+O(n^{-5/3}). \end{eqnarray} $$ So $\sqrt[3]{n^2}(\sqrt[3]{n+1}-\sqrt[3]{n}) = 1/3 + O(n^{-1}) \rightarrow 1/3$ as $n \rightarrow\infty$.