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For $\lim_{(x, y) \to (0, 0)} \frac{\sin xy}{y}$, why is it equivalent to $0$? I can't really figure out how to do squeeze theorem in this scenario, and I also can't use substitution?

Gary
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EAO123
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    $$\left|\frac{\sin(xy)}{y}\right| = |x|\left|\frac{\sin(xy)}{xy}\right|\xrightarrow{(x,y)\to(0,0)} 0\times 1 = 0$$ – Átila Correia Feb 04 '24 at 23:25
  • i hate myself, and am an idiot, thank you – EAO123 Feb 04 '24 at 23:32
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    I think it could be proven similarly like $\lim_{x\to 0}\frac{\sin x}{x}$ by squeeze theorem. Also, it is direct via polar coords. substitution: $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{y}=\lim_{r\to 0}\frac{\sin(r^2\cos\theta\sin\theta)}{r\sin\theta}\sim \lim_{r\to 0}\frac{r^2\sin\theta\cos\theta}{r\sin\theta}\to 0$$ – Joan S. Guillamet F. Feb 04 '24 at 23:35

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