Answer
$f(x)=|x|$ is continuous on $R^n$
Work Step by Step
Let us consider $|x-a|^2=(x-a) \cdot (x-a)$ ...(1)
We need to use the definition of a limit and continuity when $f$ is continuous.
We know that for a smaller value $\alpha \gt 0$, there must exist $\beta \gt 0$ (a smaller value than $\alpha$ ) such that: $|x-a| \lt \beta$ and $|f(x)-f(a)| \lt \alpha$
From equation (1), we have
$||x||-||a|| \lt |x-a|$
and $||x||-||a|| \lt |x-a| \lt \alpha $
Thus, $||x||-||a|| \lt \alpha $
or, $\alpha =\beta$
Therefore, $\lim\limits_{x \to a}f(x)=f(a)$
This shows that $f(x)=|x|$ is continuous on $R^n$
