<s>[INST] $f(x) = x^\frac{1}{3}$ is continuous at any $p \neq 0$. Trying to show $f(x) = x^\frac{1}{3}$ is continuous at any $p \neq 0$. The case with 0 is easy, this one has me stuck. I should use that $a^3 - b^3 = (a-b)(a^2+ab+b^2).$ I started off by $$|x^\frac{1}{3} - p^\frac{1}{3}| = |x^\frac{1}{3} - p^\frac{1}{3}| \frac{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|}{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|} = \frac{|x-p|}{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|}.$$ The numerator is exactly what I want it to be. The denominator is giving me headache because $(xp)^\frac{1}{3}$ can be negative, so i can't do $\leq \frac{|x-p|}{|p^\frac{2}{3}|}$ and get the $\delta$ easily. I have already shown that either $x^\frac{2}{3}+(xp)^\frac{1}{3} = x^\frac{1}{3}(x^\frac{1}{3} + p^\frac{1}{3}) \geq 0$ or $p^\frac{2}{3}+(xp)^\frac{1}{3} = p^\frac{1}{3}(x^\frac{1}{3} + p^\frac{1}{3}) \geq 0$ (or both). The first case allows me to proceed to $\leq \frac{|x-p|}{|p^\frac{2}{3}|}$, but the second leaves me with $\leq \frac{|x-p|}{|x^\frac{2}{3}|}$, so I can't do anything about the variable quantity. Is there an easier way to solve the problem? I would like not to go case-by-case, but that might be the option.
 [/INST] If you make sure that $|x-p|<|p|$ (as part of how you choose $\delta$), then $x$ and $p$ have the same sign, and all terms in the denominator are positive. 
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