Prove that the function $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=x^\frac{1}{p}$ is continuous
Here, I am trying to prove by sequential criterion.
Suppose $f:(0,\infty)\to \mathbb{R}$ by $f(x)=x^\frac{1}{p}$
I need to prove that if $x_n\to a$ then $f(x_n)\to f(a)$
I am choosing $y_n=f(x_n)=x_n^\frac{1}{p}$ and $b=f(a)=a^\frac{1}{n}$
Now, I consider $y_n^p-b^p=(y_n-b)(y_n^{p-1}+........+b^{p-1})$
$\Rightarrow |y_n-b|\le \frac{|y_n^p-b^p|}{|b^{p-1}|}\\\Rightarrow |f(x_n)-f(a)|\le \frac{|x_n-a|}{b^{p-1}}<\epsilon$
as $x_n\to a$
So, I proved for positive real numbers. How can I prove for negative real numbers? For example, for $x=-1$?
The function $f(x)=\sqrt[n]{x}$ only is defined on $[0,\infty)$ for even $n$. For odd $n$, write $f(x)=\sqrt[n]{x}$ where $x>0$ and $f(x)=-\sqrt[n]{-x}$ where $x<0$.