I need to prove the following limit and I would like to have some feedback about my thought process as I'm still a beginner.
$$\lim_{(x,y)\to (0,0)} \frac{\sin^2(xy)}{x^2+y^2}=0$$
My proof:
I use polar coordinates and I get an indeterminate form $[\frac{0}{0}]$
$$ \lim_{\rho \to 0} \frac{\sin^2(\rho^2\cos(\theta)\sin(\theta))}{\rho^2} =\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho} \cdot \lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}$$
Solving
$$\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\leq\left|\frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\right|\leq\frac{\rho^2\lvert\cos(\theta)\rvert\lvert\sin(\theta)\rvert}{\rho}\leq \rho\to 0$$
regardless of $\theta$. Thank you for your help!
Best Answer
You may use the fact that $$ |\sin x|\le |x| $$ for all $x\in\mathbb R$.
Then, for $(x,y)\ne (0,0)$, $$ 0\le \frac{\sin^2(xy)}{x^2+y^2}=\frac{\sin^2(r^2\cos\vartheta\sin\vartheta)}{r^2}\le \frac{(r^2\cos\vartheta\sin\vartheta)^2}{r^2}=r^2\cos^2\vartheta\sin^2\vartheta\le r^2\to 0 $$ as $(x,y)\to (0,0)$.