# Study the continuity of $\frac{\sin(xy)}{{xy}}$

It is needed to study the function $f(x,y)$ defined as follows $$f(x,y) = \begin{cases} \frac{\sin(xy)}{{xy}}, & \text{if }xy\text{ \neq 0} \\ 1 & \text{if }xy\text{ = 0} \end{cases}$$

I have proved that

$$\lim_{(x, y) \rightarrow (0, 0)} \frac{\sin xy}{xy}=1$$

How can I continue on defining $f$ continuity as I am not sure if the previous limit can apply to $$\lim_{(x, y) \rightarrow (a, 0)} \frac{\sin xy}{xy}=1 \text{ or } \lim_{(x, y) \rightarrow (0, a)} \frac{\sin xy}{xy}=1$$ where $a { \neq 0}$ is a real number?

• Since $a\neq0$ you your limit for $x\rightarrow a$ case is $lim_(y\rightarrow0)\frac{sin(ay)}{ay}$. And this should be easy to compute with l'Hopital. Same for $y$ case. Commented Aug 26, 2017 at 20:54

If $xy \neq 0$ then $x \neq 0$ and $y \neq 0$ so $\frac{\sin{xy}}{xy}$ is continuous as a product of continuous functions.
Now if $p=(x_0,0)$ where $x_0\neq 0$, then $(x,y) \rightarrow (x_0,0) \Rightarrow xy \rightarrow 0x_0=0$
With this you can compute the limit as $xy \rightarrow 0$ which is $1$ and prove continuity on the $x-$axis
Apply the same argument to points $(0,y_0)$ where $y_0 \neq 0$
and for $p=(0,0)$ tp show that $f$ is continuous on $y-$axis and on the origin respectively.