## Thomas' Calculus 13th Edition

Yes ; $\lim\limits_{(x,y) \to (0,0) } y \sin \dfrac{1}{x} =0$
Given : $|\sin \dfrac{1}{x}| \leq 1, |y \sin \dfrac{1}{x}| \leq y$ Re-arrange as: $0 \leq |y \sin \dfrac{1}{x}| \leq |y|$ Since, by the Squeeze Theorem $\lim\limits_{(x,y) \to (0,0) } |y \sin \dfrac{1}{x}| =0$ and $\lim\limits_{(x,y) \to (0,0) } y \sin (1/x)=0$ This implies that the limit for $\lim\limits_{(x,y) \to (0,0) } y \sin \dfrac{1}{x} =0$ by the Squeeze Theorem. So, our answer is Yes.