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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.

2 Answers 2

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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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Note that $x^{1/3}$ is the inverse of $x^3$ and the latter has nonzero derivative everywhere but at $0$.

Then, if you haven't already done so, prove as a lemma that the inverse of a differentiable function is continuous at every point where the derivative is nonzero.

This will free you from the detailed algebra and allow you to focus on the qualitative aspects of the situation.

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    The inverse of a differentiable function is even differentiable at every point where the derivative is nonzero. The inverse of an invertible continuous function on $\mathbb R$ is continuous.2012-02-10