Answer
The limit is
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1.$$
Work Step by Step
We have a given limit
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}.$$
We well use the polar coordinates:
$$\rho^2 = x^2+y^2,\quad \tan\phi=\frac{y}{x}.$$
This yields
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{\rho\to0}\frac{\sin \rho^2}{\rho^2},$$
where we used that when $(x,y)\to0$ then $\rho\to0$. When $\rho\to0$ both the numerator and the denominator tend to zero ($\rho\to0$ then $\sin\rho^2\to0$ and $\rho^2\to0$), so we can use L'Hopital rule:
$$\lim_{\rho\to0}\frac{\sin\rho^2}{\rho^2}=\lim_{\rho\to0}\frac{(\sin\rho^2)'}{(\rho^2)'} = \lim_{\rho\to0}\frac{2\rho\cos \rho}{2\rho}=\lim_{\rho\to0}\cos\rho= 1,$$
so the original limit is
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1.$$
