(Uniform) continuity in $\mathbb{R^n}$ I have the following function
$$
\mathbb{R^n} \setminus \{0\} \to \mathbb{R^n} : f(x) = \frac{x}{|x|^2}
$$
equipped with the euclidean norm (p-norm with $p = 2$)
I know I can analyze the continuity of multivariate functions by splitting it up into component functions.
So I do know that $x \mapsto x$ continous and that every norm is continuous in a metric space. 
Thus the division of two continuous functions is continuous if the denominator doesn't become $0$.
But by definition we excluded $\{0\}$ from the domain.
And what about uniform continuity? Can I somehow extend the $\epsilon-\delta$ definition from $\mathbb{R}$ to $\mathbb{R^n}$?
 A: You can do the same thing: A function $F=(f_1,\dots,f_n):\mathbb{R}^n\to \mathbb{R}^n$ is uniformly continuous if its component functions are. 
Suppose $f_1,\dots,f_n$ are uniformly continuous functions from $\mathbb{R}^n\to \mathbb{R}$. Fix $\epsilon>0$ and let $\delta=\min_{1\leq i\leq n}\delta_i$ be the minimum of the $\delta_i$ meeting the $\sqrt{\epsilon/n}$ challenge for each $f_i$ and any points $x,y\in \mathbb{R}^n$. 
Then, for this fixed $\epsilon>0$ and any $x,y\in \mathbb{R}^n$, we have  $|x-y|\leq \delta$ implies
$$
|f_i(x)-f_i(y)|<\epsilon, 1\leq i\leq n. 
$$ 
So,
$$
||F(x)-F(y)||^2=\sum_{i=1}^n |f_i(x)-f_i(y)|^2< \sum_{i=1}^n \epsilon/n=\epsilon
$$
It's worth internalizing the mantra, vector valued functions inherit the properties of their component functions, (within reason).
Details on this particular example: If we take $\epsilon=1$, we may choose for any given $\delta>0$, the points $x=(\delta,0,\dots,0), y=(\delta/2,0,\dots,0)$. Then, $||x-y||=\delta/2<\delta$ but 
$$
||f_1(x)-f_1(y)||=\left|\frac{x_1}{|x|^2}-\frac{y_1}{|y|^2} \right|=\frac{1}{\delta}\geq 1
$$
for $\delta\leq 1$. I leave $\delta>1$ to you. 
A: Restricted to $\mathbb{R} \times \{0\}^{n-1}$, $f$ si just the usual inverse function $f(x) = \frac{1}{x}$, which is not uniformly continuous. So $f$ cannot be.
Uniform continuity is a metric statement : you just replace $|.|$ by a distance ($d(x,y) < \delta$) and in that case by the norm distance $\|x-y \|$.
A: If $D\subset\Bbb R$ and $f:D\to \Bbb R$ and if $f$ is unbounded on some bounded $E\subset D$ then $f$ is not uniformly continuous. In the proof of this, just replace $\Bbb R$ with $\Bbb R^n$ and replace $|u|$ with $\|u\|$ for any $u\in \Bbb R^n.$
In your Q let $E=\{x\in \Bbb R^n: 0\ne\|x\|<1\}.$
