<s>[INST] Show that $x^\frac{1}{n}$ is continuous at all $a \in [0,\infty)$ I have to show $x^\frac{1}{n}$ is continuous using:
$|x-a| = |(x^\frac{1}{n})^n - (a^\frac{1}{n})^n| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle \sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k} |$
Here's what I did:
$a=0$:
$|x^\frac{1}{n}| < \epsilon => |x| < \epsilon^n$
Pick $\delta = \epsilon^n$, and we have $f(x)$ is continuous at $x=0$
$a>0$:
$|x-a| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}| < \epsilon \cdot |\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}|$
Since this sum evaluates to a real number $R > 0$, we can pick $\delta = R\epsilon $  and we get:
$0<|x-a| < \delta => |x^\frac{1}{n} - a^\frac{1}{n}| < \epsilon$
Is this correct or is there anything else I need to show?
Also, this is my first time formatting with MathJax, so if there are any errors please let me know!
 [/INST] First you need the correct factorization
$$x- a = (x^{\frac{1}{n}} - a^{\frac{1}{n}})\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}}$$
Whence,
$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|= \frac{|x-a|}{\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|} $$
Note that by the reverse triangle inequality $|a|- |x| \leqslant |x-a|$ which implies that $|x| \geqslant |a| - |x-a|$.
If $a > 0$ and $|x-a| \leqslant \frac{a}{2}$, then we have $x = |x| \geqslant |a| - |x-a| \geqslant a - \frac{a}{2} = \frac{a}{2}.$
Hence, $\displaystyle x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \geqslant 2^{\frac{k}{n}}\left(\frac{a}{2}\right)^{1 - \frac{1}{n}},$ and
$$\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|\geqslant  \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\sum_{k=0}^{n-1}2 ^{\frac{k}{n}}  = \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\frac{1}{2^{\frac{1}{n}} -1} := C_n$$
It follows that if  $|x-a| < \delta = \min\left(\epsilon C_n, \frac{a}{2}\right)$, then
$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|\leqslant \frac{|x-a|}{C_n} < \epsilon,$$
and $x \mapsto x^{\frac{1}{n}}$
is continuous at $a > 0$.
</s>