Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$. I want to show that

$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$

Is it valid to do it like this:
$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to (0,0)} |\exp(xy)\cdot xy\cdot (x^2-y^2)|=0$$

We may restrict the neibourhood to $x^2+y^2<1$, then $e^{xy}<2$ and
$$\left|\dfrac{e^{xy}xy(x^2-y^2)}{x^2+y^2}\right|\leqslant\dfrac{2|xy|(|x|^2+|y|^2)}{x^2+y^2}\leqslant x^2+y^2$$
