Limit of a multivariable function approaching to $(0,0)$ So I have
$$\lim_{(x,y) \to (0,0)}\frac{x\sin(xy)}{y}$$
And I was wondering if I'm allowed to do this:
$$= \lim_{x \to 0} x * \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{y} \\=0 * \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{y} \\ = 0$$
 A: \begin{align*}\lim_{(x,y) \to (0,0)}\frac{x\sin\ (xy)}{y}&= \lim_{(x,y) \to (0,0)}\  x^2\bigg(\frac{\sin\ (xy)}{xy}\bigg)\\&=\lim_{x \to 0}\ x^2 \cdot \lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{xy} \\&=0 \cdot 1=0 \end{align*}
A: Note that for $x\neq 0$ we have $\frac{x}{x}=1$, so we can write $$\lim_{\left(x,y\right)→ \left(0,0\right)} \dfrac{\sin  xy}{y}= \lim_{\left(x,y\right)→ \left(0,0\right)} \dfrac{x\ \sin xy}{xy}=\lim_{\left(x,y\right)→ \left(0,0\right)} x\dfrac{\sin xy}{xy} $$ and thus if $\lim_{\left(x,y\right)→ \left(0,0\right)} \dfrac{\ \sin xy}{xy}$ exists then we have
$$\lim_{\left(x,y\right)→ \left(0,0\right)}\dfrac{\ \sin xy}{y}=\left(\lim_{\left(x,y\right)→ \left(0,0\right)}x\right)\left(\lim_{\left(x,y\right)→ \left(0,0\right)}\dfrac{\ \sin xy}{xy}\right)=0.\left(\lim_{\left(x,y\right)→ \left(0,0\right)}\right)\frac{\ \sin xy}{xy}=0 $$
But let $t=xy$. Then we have
$$\lim_{\left(x,y\right)→ \left(0,0\right)} \dfrac{\sin xy}{xy}=\lim_{t→ 0}\dfrac{\ \sin t}{t} $$
and we know tha $\lim_{t→ 0}\dfrac{\sin t}{t}=1$ so the limit exists
and we have $$\lim_{\left(x,y\right)→ \left(0,0\right)} \dfrac{\sin  xy}{y}=0$$
