Show $f(x):=x^p$, where $p \ge 0$, is continuous For p being an integer, the result follows immediately from the Algebraic Limit Theorem and Sequential Criterion. If p is not an integer, however, I cannot use these results. I have tried solving from first principles given $|x-a|<\delta$ but I can't see how to simplify $|x^p-a^p|$. Any hints?
 A: You know that the function $x \mapsto e^x$ is continuous on $\mathbb{R}$. Hence for $a \in \mathbb{R}^{*+}$, for all $\epsilon>0$, it exists $\eta>0$ so that $\left|x-a\right|<\eta$ implies $\left|e^{x}-e^{a}\right|<\epsilon$.
Then 
$$
\left|x^{p}-a^{p}\right|=\left|e^{p \ln\left(x\right)}-e^{p \ln\left(a\right)}\right|<\epsilon
$$
A: A bit more:
For $x>0$, the function 
a)$\exp(x)$, defined and continous on $\mathbb{R}$.
b) $\log(x)$, defined and continous on $\mathbb{R^+_{\star}}.$
Consider  $p>0$, given.
$f(x) = x^p := \exp(p\log(x)).$
$f$ being the composition of 2 continuous functions is continuous on $ \mathbb{R^+_{\star}}.$
The problem :
Show that $f(x)=x^p$ is continuous at $x= a >0.$
$\epsilon,$ $ \delta$ proof.
Plan:
1) Show that $p\log(x)$ is continous at $x=a.$
2) Show that $\exp(y)$ is continous at $y= p\log(a)$.
3) Along the lines of user layman's comment show that the composition of the $2$ functions is continuos.
Note :$\exp$ and $\log $ are base $e$ above.
Comments welcome.
