# finding the limit of a sequence

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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 i don't think it matters for this problem but the limit goes to +infinity – nofe Dec 17 '11 at 16:35 use difference of cubes $a^3-b^3=(a-b)(a^2+ab+b^2)$ to simplify the expression – yoyo Dec 17 '11 at 16:41

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

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

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 thanks, this is what i'm looking for. I don't want to use l'hospital or differentiate anything. How did you know this was the adjacent? how do you find the adjacent in a polynomial higher than 2? – nofe Dec 18 '11 at 0:16 @nofeliram,because of third root terms it is obvious that you should use difference of cubes formula... – pedja Dec 18 '11 at 8:25 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. – nofe Dec 18 '11 at 8:56

\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 $n \rightarrow\infty$.

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