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With the change of variables, we have $\displaystyle\lim_{(x,y) \to (0, a)} \frac{\sin(xy)}x = \displaystyle\lim_{(x,y) \to (0, 0)} \frac{\sin(x(y+a))}x$

Now we perform the exchange of variables $\begin{cases} x = r \cos \theta \\ y = r \sin \theta\end{cases}$ and get

$\displaystyle\lim_{(x,y) \to (0, a)} \frac{\sin(xy)}x = \displaystyle\lim_{r \to 0} \frac{\sin(r^2 \cos \theta \sin \theta + ar \cos\theta)}{r \cos \theta}$.

The Taylor expansion for $\sin(r^2 \cos \theta \sin \theta + ar \cos\theta)$ in $r$ at $r = 0$ is $\sin(r^2 \cos \theta \sin \theta + ar \cos\theta) = ar \cos \theta + \mathcal{O}(r^2).$

If one now tries to take the limit, one will have

$\displaystyle\lim_{r \to 0} \frac{\sin(r^2 \cos \theta \sin \theta + ar \cos\theta)}{r \cos \theta} = \displaystyle\lim_{r \to 0} \frac{ar \cos \theta + \mathcal{O}(r^2)}{r \cos \theta} = \displaystyle\lim_{r \to 0} a + \mathcal{O}(r) = a.$

According to this reasoning, the limit should exist for all $a$, and be equal to $a.$ Is this, however, a valid proof for the limit? I tried an $(\varepsilon, \delta_\varepsilon)$-approach as well, but I didn't get anything "good" to work with.

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    One thing to note is that the function is not defined anywhere on the $y$-axis. In what sense can the limit exist in this case? – 2017-01-15
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    @zhw. The definition is very similar to the regular epsilon-delta definition, but here you let the abs() function denote the Euclidean distance in $R^2$. So we say that f(x ,y) approaches L as (x, y) approaches (a,b) if and only if∀ε>0∃δ> : (∀(x, y) in D) |(x, y) - (a,b) | < δ implies |f(x, y) - L| < ε. – 2017-02-05

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HINT: write $$y\cdot \frac{\sin(xy)}{xy}$$ and the searched limit is $a$