Why is $\sin(xy)/y$ continuous? Me and my mates are crunching this question for a while now.
While we know that $\sin(xy)$ is continuous , $1 / y $ as the other part of the function clearly has a continuity gap at $y = 0 $, though the function can be continued at $y = 0$ with $f(x,0) = 0 $- why is that?
We tried some things but are not getting to the important step that proves the matter.
 A: Recall that for $z$ real, $|sin (z)| \leq |z|$.
A: I suppose the question is why the function is continuous for $x,y$ in $\mathbb{R}$ ...
The function is obviously continuous in every $(x,y)$, where $x$ and $y$ are non-zero.
Let's study the $3$ cases, when $x=0$, $y=0$, and $(x,y)=(0,0)$
1.Let $x$ be any non-zero Real number:
$\frac{\sin(xy)}{y} = \frac{\sin(xy)}{xy}x \to  1\times x$ when $y\to0$, so the function is continous and the limit is $x$


*

*Let y be any non-zero Real number:
$\frac{\sin(xy)}{y} = 0$ when $x=0$, obvious continuous

*When $(x,y)\to(0,0)$, from $1$ and $2$ above, the function has a limit, $0$.
A: You can replace $f(x, 0) = x$. Let's fix $x = a\neq 0$. Then the function essentially becomes $g(y) = \frac{\sin (ay)}{y}$. To figure out the limit for "$g(0)$", we can substitute $z = ay$:
$$
\lim_{y \to 0}f(a, y) = \lim_{z \to 0}\frac{\sin z}{z/a} = a\cdot\lim_{z \to 0}\frac{\sin z}{z} = a
$$
For $x = 0$, we have that $f(0, y) = 0$ for non-zero $y$, so $f(0, 0) = 0$ is a natural extension at the origin as well.
A: This seems to be wrong if the question is whether the function
$$
f:\mathbb{R}^2 \to \mathbb{R}:(x,y) \mapsto \frac{\sin(xy)}{y},
$$
extended by $0$ where $y=0$ is continuous. Take any point $(a,0)$. Then $(a,1/n)$ is a sequence that converges to this point, so by continuity $f(a,1/n)$ should converge to $f(a,0)=0$. However
$$
\frac{\sin(a/n)}{1/n}=\frac{\sin(a/n)}{a/n} \cdot a \to a,
$$
so if anywhere, it can only be continuous in the origin. 
A: The expression
$$f(x,y):={\sin(xy)\over y}$$
is at face value undefined when $y=0$, but wait: When $y\ne 0$ one has the identity
$${\sin(xy)\over y}=\int_0^x\cos(t\>y)\ dt\ .$$
Here the right side is obviously a continuous function of $x$ and $y$ in all of ${\mathbb R}^2$. It follows that the given $f$ can be extended continuously to the full plane in a unique way.
