Is the following function continuous ?
f : $\mathbb{R^3}\times \mathbb{R^3} \rightarrow \mathbb{R^3} $
$x,y \mapsto x \times \frac{y-x}{|y-x|}$ for $x \neq y$, else $0$.
Now I would have argued that for the potential discontinuity at the points $x_0=y_0$, continuity holds since :
$ \lim_{x,y \rightarrow x_0,y_0} |f(x,y)| = \lim_{x,y \rightarrow x_0,y_0} |x| | \frac{y-x}{|y-x|} ||\sin(\theta)|= \lim_{x,y \rightarrow x_0,y_0} |x| |\sin(\theta)| \rightarrow 0 $
Is that true or have I missed a point ?
In your argument, you write
$$\lim_{x,y \rightarrow x_0,y_0} |x| |\sin(\theta)| \rightarrow 0$$
It seems that you assume $\sin(\theta) \rightarrow 0$ as $x \rightarrow x_0$, where $\theta$ is an angle between $x$ and $y-x$ (or $\frac{y-x}{|y-x|}$). But how do you know $\sin(\theta) \rightarrow 0$?
Well, $f$ is not continuous in fact. For example, let $\{e_1, e_2, e_3\}$ be a standard basis for $\mathbb R^3$ and consider a path $x(t)=e_1$, $y(t)=e_1+te_2$ $(t \in \mathbb R)$. Then $y-x=te_2$. Thus $f(x, y)= \begin{cases}e_1 \times e_2=e_3 & \text{if } t>0 \\ e_1 \times (-e_2)=-e_3 & \text{if } t<0 \end{cases} $
Note that $y \rightarrow x=e_1$ as $t \rightarrow 0$. $$f(e_1, e_1)=0$$$$\lim_{t \to 0+}f(x(t), y(t))=e_3$$$$\lim_{t \to 0-}f(x(t), y(t))=-e_3$$
It follows that $f$ is not continuous at $(e_1, e_1)$. It is also recommended to compute $|\sin \theta|$ in this example.