Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \sin \sqrt {{x^2} + {y^2}} \cr
& {\text{Rewrite the limit using polar coordinates}} \cr
& x = r\cos \theta ,{\text{ }}y = r\sin \theta \cr
& \left( {x,y} \right) \to \left( {0,0} \right),{\text{so }}r \to 0 \cr
& {\text{Substituting}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \sin \sqrt {{x^2} + {y^2}} = \mathop {\lim }\limits_{r \to 0} \sin \sqrt {{{\left( {r\cos \theta } \right)}^2} + {{\left( {r\sin \theta } \right)}^2}} \cr
& = \mathop {\lim }\limits_{r \to 0} \sin \sqrt {{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} \cr
& = \mathop {\lim }\limits_{r \to 0} \sin r \cr
& {\text{Evaluate the limit when }}r \to 0 \cr
& = \sin \left( 0 \right) \cr
& = 0 \cr
& {\text{Then, we can conclude that}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \sin \sqrt {{x^2} + {y^2}} = 0 \cr} $$
