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

Solution 1:

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.

Solution 2:

$\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{ n \left[ \left( 1 + \frac 1n \right)^{\frac 13} - \left( 1 + \frac 1{n^3} \right)^{\frac 13} \right] \right\} = \\ \displaystyle \lim_{n \rightarrow \infty} \left[ n \left( 1 + \frac 1{3n} - 1 - \frac 1{3n^3} \right) \right] = \lim_{n \rightarrow \infty} \left( \frac 13 - \frac 1{3n^2} \right) = \frac 13$