Given the function $f: R^{n+1}- \{0\} \rightarrow S^n$ defined by $f(x)= \dfrac{x}{\Vert x\Vert}$, prove that $f(x)$ is continuous.
I have tried to use the fact that $\mathbb{R}^{n+1}-\{0\}, S^n$ are metric spaces, so that I can use an epsilon-delta proof.
However, I was not able to get to the last part of the inequality
$$ \left\Vert\frac{x}{\Vert x\Vert}-\frac{x_0}{\Vert x_0\Vert}\right\Vert.$$
I am open to another method.
Scalar multiplication is continuous as a map $s: \Bbb R \times \Bbb R^{n+1} \to \Bbb R^{n+1}, s(t,x)=t \cdot x$. This follows by inequalities like
$\|t\cdot x - t'\cdot x'\| =\|(t\cdot x - t'\cdot x) + (t' \cdot x - t' \cdot x)\| \le |t-t'| \cdot \|x\| + |t'|\|x- x'\|$, from which we can show that whenever $t_n \to t_0$ in $\Bbb R$ and $x_n \to x_0$ in $\Bbb R^{n+1}$ we also have $t_n \cdot x_n \to t_0 \cdot x_0$ in $\Bbb R^{n+1}$ as well.
Also the norm map is continuous from $\Bbb R^{n+1}$ to $\Bbb R$ so if $x_n \to x_0$ in $\Bbb R^{n+1}\setminus \{0\}$, we have that $\|x_n\| \to \|x_0\|$ and as $g(x)=\frac{1}{x}$ is continuous from $(0,\infty)$ to $\Bbb R$, we also have $\frac{1}{\|x_n\|} \to \frac{1}{\|x_0\|}$ and then applying the first fact about scalar multiplication, $f(x_n) = \frac{x_n}{\|x_n\|} = \frac{1}{\|x_n\|} \cdot x_n \to \frac{1}{\|x_0\|} \cdot x_0 = \frac{x_0}{\|x_0\|} =f(x_0)$. QED.
We don't need sequences (but they suffice), and can also write $f$ as a composition of continuous maps instead. But that's the basic idea: norm, and scalar multiplication on $\Bbb R^{n+1}$ and inversion on the reals are all continuous and $f$ is a combination of these operations.