Continuity of piecewise function $$f(x,y) = \begin{cases} \dfrac{\sin(xy)}{xy} & \text{if $x y \ne 0$} \\ 1 & \text{if $xy=0$} \end{cases}$$
all ideas are appreciated
i think this is non-continuous, i did by converting to polar coordinates
Looking for more ideas and interesting observations
 A: I assume that $f(x)$ should read as $f(x,y)$. Define the function $g:\mathbb R^2\to\mathbb R$ as $g(x,y)=xy$ for all $(x,y)\in\mathbb R^2$. This function is clearly continuous. Moreover, define another function $h:\mathbb R\to\mathbb R$ as
\begin{align*}h(z)=\begin{cases}\dfrac{\sin z}{z}&\text{if $z\neq0$,}\\1&\text{if $z=0$.}\end{cases}\end{align*}
This function is continuous, as it is well-known that $\lim_{z\to0}(\sin z)/z=1$. Now, observe that $f=h\circ g$ and recall that the composition of continuous functions is continuous.
A: If $y$ is presumed constant (as you write $f(x)$), the function is indeed continuous wich can be proven by showing that
$$\lim_{x\to0} \frac{\sin x}x = 1$$
(or trivially if $y=0$)
And noting that $f = x\mapsto \frac{\sin x}x \circ x\mapsto xy$.
If $y$ is also a variable (i.e. you meant $f(x,y)$) take a look if $(x,y) \mapsto xy$ is continuous (yes, it is) and note again that
$$f = x\mapsto \frac{\sin x}x \circ (x,y) \mapsto xy$$
is continuous as the composition of two continuous functions.
