How to find the following limit? $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{xy}$ I am facing challenges with such type of limits.
$$\lim_{(x,y)\to(0,0)}\dfrac{\sin(xy)}{xy}$$
I tried to search on the internet but couldn't find a solution
 A: Let $f(x)=\frac{\sin x}x$. This is a continuous function everywhere except $x=0$, where the limit exists and is $1$. Thus, if we also define $f(0)=1$, we have a continuous function $\Bbb R\to\Bbb R$. Thus $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{xy}=\lim_{(x,y)\to(0,0)}f(xy)=f\left(\lim_{(x,y)\to(0,0)}xy\right)=f(0)=1.$$
Edit: As pointed out by Ted Shifrin, this function is not defined in a neighborhood of $(0,0)$ because it is undefined on the lines $x=0$ and $y=0$. Thus this becomes a definitional issue - some definitions require the function to be defined for $(x,y)\ne(0,0)$, others only require it to be close to the putative limit on its points of definition. Under the first interpretation, this limit does not exist, and under the second interpretation, it is $1$ because $\frac{\sin x}x=f(x)$ whenever it is defined, and $f(xy)$ does indeed have limit $1$.
A: draw a graph of a sector of circle with centre $O$, centre angle $x$.
we will find that: $$\sin(x)<x \implies \sin(x)/x < 1$$ and $$x < \tan(x) = \sin(x)/\cos(x)$$
then we have $$\cos(x) < \sin(x)/x < 1.$$
By squeeze theorem,  $\lim \sin(x)/x =1$ as $x$ tends to $0^+$.
A: If you don't want to see the denominator, you can write: $\sin(xy) = xy - \dfrac{(xy)^3}{3!} + \dfrac{(xy)^5}{5!} -....+$, and divide both sides by $xy$ to get:
$\dfrac{\sin(xy)}{xy} = 1 + o((xy)^2)$, from this you get the limit is $1$.
