# Prove that $f(x) = x^{1/5}$ is continuous everywhere

Need to prove that $f(x) = x^{1/5}$ is continuous everywhere, where $f: \mathbb{R} \to \mathbb{R}$:

from definition we need to show that given $\epsilon > 0$ $\exists \delta > 0$ s.t. $|x-x_0|<\delta \Rightarrow \left|x^{\frac{1}{5}} - x_0^{\frac{1}{5}}\right| < \epsilon$ for any point $x_0 \in \mathbb{R}$

I have a proof but it's somewhat unjustified:

consider $\left|x^\frac15 - x_0^\frac15\right| \geq \left|x^\frac15\right| - \left|x_0^\frac15\right|$ from the triangle inequality since $\left|x^\frac15\right| < |x|$ and $\left|x_0^\frac15\right| < |x_0|$ then $\left|x^\frac15 - x_0^\frac15\right| \geq \left|x^\frac15\right| - \left|x_0^\frac15\right| < |x| - \left|x_0\right| = \left|x-x_0\right| = \delta$ so we can choose $\delta = \epsilon$?

Overall I'm not happy with the proof, in the last inequality I don't think I can just state that delta = epsilon and be done, but I have no idea what else to do. I also am not sure about this step $|x| - |x_0| = |x-x_0|$and $\left|x^\frac15\right| < |x|$ and $\left|x_0^\frac15\right| < |x_0|$ that step also...

if anyone could help me out.. thank you

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Careful: $|x|-|x_0| \neq |x-x_0|$ . Take, e.g., $x<x_0$ , to see what happens. – user99680 Dec 7 '13 at 19:39
Thank you.. any ideas on what to do with the proof? I still cannot proceed – user113494 Dec 7 '13 at 20:18
Let me give it a try; I'll be back in a while. – user99680 Dec 7 '13 at 20:19
$x \rightarrow x^5$ is continuous (since it's a polynomial) and bijective. So its inverse is also continuous. See for example A continuous bijection f:R→R is an homeomorphism? – Jean-Claude Arbaut Dec 8 '13 at 1:56
Or, you can use that both $e^x$ and $lnx$ are continuous, and $x^{1/5} =e^{(1/5)lnx}$ is the composition of the two continuous functions $f(x)=(1/5)lnx$ and $g(x)=e^x$, and therefore it is continuous. I wrote it below. – user99680 Dec 8 '13 at 6:15

If you accept that both $e^x$ and $lnx$ are continuous, then, working with $x$ in $(0, \infty)$ you can do this:

$x^{1/5}=e^{(1/5)lnx}$ , is the composition of the two continuous functions $e^x$ and $\frac{1}{5}lnx$ , so, as the composition of two continuous functions, it is continuous.

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It is not true in general that $|x|^{\frac{1}{5}}<|x|$ (this is true iff $|x|>1$).

Here's how I would go about a proof. First treat continuity at $0$ as a separate case ($\delta=\varepsilon^{5}$). Now let $x\in\mathbb{R}$ and $\varepsilon>0$ be arbitrary. Set $\delta=\min\{\varepsilon|x|^{\frac{4}{5}},|x|\}$.

We need to do a bit before we finish the proof. Notice that for $x,y\in\mathbb{R}$ such that $xy>0$ we have $x^{5}-y^{5}=(x-y)(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4})$ so $$|x-y|=\frac{|x^{5}-y^{5}|}{|x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}|}=\frac{|x^{5}-y^{5}|}{|x^{4}|+|x^{3}y|+|x^{2}y^{2}|+|xy^{3}|+|y^{4}|} \leq\frac{|x^{5}-y^{5}|}{|x|^{4}}$$ The above string of inequalities implies $$\textbf{(1) }\text{for } x,y\in\mathbb{R}\text{ with } xy>0, |x^{\frac{1}{5}}-y^{\frac{1}{5}}|\leq\frac{|x-y|}{|x|^{\frac{4}{5}}}$$

Now let $|x-y|<\delta$. Then by (1) we have $$|x^{\frac{1}{5}}-y^{\frac{1}{5}}|\leq\frac{|x-y|}{|x|^{\frac{4}{5}}}<\frac{\delta}{|x|^{\frac{4}{5}}}\leq\varepsilon$$ Then we are done. We need $\delta<|x|$ so that $xy>0$. Why do we need $xy>0$? I'll leave that bit to you.

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The derivations in the first line, where $|x-y|$, is not correct. – Mhenni Benghorbal Dec 8 '13 at 5:40
What's wrong with them. They are made with the assumption that xy>0. – Eric Dec 8 '13 at 18:25