Master limit 1

By changing to polar coordinates we can evaluate this.

Making the subs: $x=r\cos(\theta),\,y=r\sin(\theta)\implies x^2+y^2=r^2$

Which changes the limit to: $\lim_{r\to0}\frac{\cos(\theta)\sin(r\sin(\theta))-\sin(\theta)\sin(r\cos(\theta))}{r}$

Using L'Hôpital's rule we find:

$\lim_{r\to0}\frac{\cos(\theta)\sin(\theta)\left(\cos(r\sin(\theta))-\cos(r\cos(\theta))\right)}{1}= 0$

Jun!!!. It wasn't solve in this way. Follow Isaac solution or you can solve it also by using inequalities

Intuition

After imagining the overall behavior of the function by imagining the values at extreme values, at special cases $y=\pm x$ and varying one variable while the other is a constant. I immediately became convinced it is 0.

To prove it

I split the limit into two terms $\lim_{(x,y)\to (0,0)} \frac{\sin (y)}{x+y^2/x}$ regarded $x$ as a constant taking the limit as $y \to 0$ then it's always zero. The other term has the same result by analogy and symmetry. Therefore the limit is always zero.

Improvement and corrections on my thought processes are always appreciated.

obviously it is zero since (x,y) has the same value..