# Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$

$$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$

I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$

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Note $n^3+n^2$ is roughly the cube of $n+(1/3)$. – Gerry Myerson Nov 14 '12 at 2:49
Try writing $\sqrt[3]{n^3+n^2} = n\sqrt[3]{1+1/n}$ then using the power series for $\sqrt[3]{1+x}$. – Antonio Vargas Nov 14 '12 at 2:51

You should use that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$, $b=\sqrt[3]{n^3+1}$ and then multiply your expression by $(a^2+ab+b^2)/(a^2+ab+b^2)$. Then use the trick you are trying to use.
You could (and maybe you should) replace You should by You could. Consider for example the analogous quantity $\sqrt[\pi]{n^\pi+n^{\pi-1}}-\sqrt[\pi]{n^\pi+1}$, then your approach fails while others yield immediately the limit $1/\pi$. – Did Nov 14 '12 at 9:49
$\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{ n \left[ \left( 1 + \frac 1n \right)^{\frac 13} - \left( 1 + \frac 1{n^3} \right)^{\frac 13} \right] \right\} = \\ \displaystyle \lim_{n \rightarrow \infty} \left[ n \left( 1 + \frac 1{3n} - 1 - \frac 1{3n^3} \right) \right] = \lim_{n \rightarrow \infty} \left( \frac 13 - \frac 1{3n^2} \right) = \frac 13$
$\displaystyle \lim_{x \rightarrow \infty} \left( 1 + \frac 1n \right)^{\frac 13} = \lim_{x \rightarrow \infty} \left( 1 + \frac 1{3n} + O\left( \frac 1{n^2} \right) \right) = \\ \displaystyle \lim_{x \rightarrow \infty} \left( 1 + \frac 1{3n} \right) + \lim_{x \rightarrow \infty} \left[ O \left( \frac 1{n^2} \right) \right] = \lim_{x \rightarrow \infty} \left( 1 + \frac 1{3n} \right)$ – Kaster Nov 14 '12 at 5:08