Mathematics - Limits Question with Solution | TestHub

By the binomial expansion,

$$\sqrt[3]{x\pm 1}=x^{1/3}\left(1\pm \frac{1}{3x}-\frac{1}{9x^2}+O(x^{-3})\right).$$

The odd terms cancel, so the parenthesis is $-\frac{2}{9}{x}^{-5/3}+O(x^{-11/3})$. Consequently $p=5/3$, $L=-2/9$, and $p^2/L=(25/9)/(-2/9)=-25/2=-12.50$.