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
 A: 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.
A: 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.
