Proving $\lim_{(x,y)\to(0,0)}\frac{\sin^2(xy)}{x^2+y^2}=0$

Question

I tried transforming cartesian coordinates to polar coordinates, which yields $\lim_{\rho\to0}\frac{\sin^2(\rho^2\sin{\theta}\cos{\theta})}{\rho^2}$ but can't proceed any further. I can't find a function $g(\rho)$ such that $|f(\rho, \theta) - L| < g(\rho)$ when $\rho\to0$. I'm pretty sure this will turn out to be easy, even trivial (it's listed as one of the first exercises of my Analysis II book). Any suggestions?

Answer 1

As $\sin t$ is asymptotic to $t$, you may consider

$$\frac{x^2y^2}{x^2+y^2}=\frac1{\dfrac1{x^2}+\dfrac1{y^2}}\to0.$$

Answer 2

We have that

$$\frac{\sin^2(xy)}{x^2+y^2}=\frac{\sin^2(xy)}{(xy)^2}\frac{x^2y^2}{x^2+y^2} \to 1\cdot0=0$$

indeed (by polar coordinates, as an alternative):

$$\frac{x^2y^2}{x^2+y^2}=r^2\cos^2\theta\sin^2\theta \to 0$$

or as an alternative we can also use that $x^2y^2\lt (x^2+y^2)(x^2+y^2)$.