I wish to prove that $\lim_{(x,y)\to(0,a)} {\sin(xy) \over x} = a$.
The ($\epsilon$,$\delta$) limit definition in this case is:
$\forall \epsilon >0, \exists \delta >0$ such that $0<\sqrt{x^2 +(y-a)^2}<\delta \Rightarrow \left|{sin(xy) \over x} - a\right| < \epsilon.$
Using the inequality $\left|{sin(a) \over a} - 1\right|\le a^2$ $(\forall a\ne 0)$, I got the following:
$\left| {sin(xy) \over x} - a\right|=\left|y\right|\left|{sin(xy)\over xy} - {a \over y}\right| \le \left|y\right|\left(\left|{sin(xy)\over xy} - 1\right| + \left|1 - {a \over y}\right|\right) \le \left|y\right| \left((xy)^2 + \left|1 - {a \over y}\right|\right) =x^2\left|y\right|^3 + \left|y - a\right|$
I suspect that I will have to choose my $\delta$ to be the minimum of some functions of $\epsilon$ and constants. What would you suggest in order to find such functions of $\epsilon$ and constants?
Hint: First look for a constant bound $|(x,y)|<1$, in particular $|x|<1$, hence $x^2<1$. Now, if $|(x,y)|<(\varepsilon /2)^{1/3}$, in particular $y<(\varepsilon /2)^{1/3}$ hence $x^2|y|^3<(\varepsilon /2)$
Can you take it from here? You're looking at $\delta = \min \left \{ 1,\varepsilon /2, ? \right \}$