Continuity of $\sqrt[3]{x}$.

Question

I have to prove that $\root{3}\of{x}$ is continuous everywhere. For this take a point $a\in\mathbb{R}$. For this to be continuous we need to find a $\delta$ for each $\epsilon$. $$|\root{3}\of x-\root 3\of a|<\epsilon$$ Find a $\delta$ for when the above holds such that $|x-a| <\delta$. $$|\root 3\of x - \root 3 \of a|\cdot|\root 3\of {x^2} + \root 3 \of{xa}+ \root 3 \of {a^2}|<\delta$$ Can't think how to proceed to get $\delta$ in terms of $\epsilon$.

Answer 1

If $|x-y| \le 1$ and $0 <a <x, y < b$, then, since $u^n-v^n =(u-v)\sum_{k=0}^{n-1} u^kv^{n-1-k} $, $x-y =(x^{1/n}-y^{1/n})\sum_{k=0}^{n-1} x^{k/n}y^{(n-1-k)/n} $ so $|x-y| \gt|x^{1/n}-y^{1/n}||\sum_{k=0}^{n-1} a^{k/n}a^{(n-1-k)/n}| =|x^{1/n}-y^{1/n}|a^{(n-1)/n} $, so that $|x^{1/n}-y^{1/n}| \lt \dfrac{|x-y|}{a^{(n-1)/n}} $.

This should be enough to get a $\delta-\epsilon$ proof.

Answer 2

simply note that $\sqrt[n]{a-b} \geq \sqrt[n]a - \sqrt[n]b$ for all $a \geq b \geq 0$ .This can be proved by considering $t = \frac{a}{b} \in (0,1]$ and $t^{1/n} \geq t$ or some other tricks.