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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) .$

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    use difference of cubes $a^3-b^3=(a-b)(a^2+ab+b^2)$ to simplify the expression2011-12-17

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

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

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    thanks! it's not obvious to me I'm just now returning to school after 4 years (army) of not touching any math at all since high school.2011-12-18
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$ \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*} $

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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 $n \rightarrow\infty$.