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I'm asked to show that the limit of $\frac{\sin(xy)}{\sqrt{x^2+y^2}} = 0$ as $(x,y) \to (0,0)$. I guess the easiest way to do this would be by converting to polar form, so this is what I did:

$$\frac{\sin(xy)}{\sqrt{x^2+y^2}} = \frac{\sin(r^2\cos\theta \sin\theta)}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}} = \frac{\sin(r^2\cos\theta \sin\theta)}{\sqrt{r^2(\cos^2\theta+\sin^2\theta)}} = \frac{\sin(r^2\cos\theta \sin\theta)}{r}$$

from here I guess I should try to prove using the definition, so I did the following:

since we know L is $0$

$$\left|\frac{\sin(r^2\cos\theta \sin\theta)}{r} - 0\right| = \left|\frac{\sin(r^2\cos\theta \sin\theta)}{r}\right| \leq \left| \frac{1}{r} \right| = \frac{1}{r}$$

which is true because $sin$ is bounded by $0$ and $1$, and $r$ is always a positive value. I also know that, since we are using polar coordinates, the distance between $(0,0)$ and $(x,y)$ is $r$. But I'm stuck here, I don't really know where to go from this. I guess I'm not getting the strategy behind the definition?

I think I'm pretty close to the answer, so feel free to give me the last steps if there's little to be done, it won't spoil the fun for me :)

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    $\begingroup$ Use $|\sin\phi|<|\phi|$ as $\phi\to0$ $\endgroup$ – Nosrati Jan 30 '17 at 4:52
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Hint: $\;\frac{\sin(xy)}{\sqrt{x^2+y^2}} = \frac{\sin(xy)}{xy} \cdot \frac{xy}{\sqrt{x^2+y^2}}\,$, then note that $\;\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \sqrt{\frac{1}{2}|xy|}\;$ by AM-GM.

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    $\begingroup$ The constant on the right hand side should be $1/\sqrt{2}$ (not $1/2$ as stated). $\endgroup$ – Fabian Jan 30 '17 at 5:38
  • $\begingroup$ @Fabian Right, thank you. Edited and fixed. $\endgroup$ – dxiv Jan 30 '17 at 5:42
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$$0\le|\frac{\sin(xy)}{\sqrt{x^2+y^2}}|=\frac{|\sin[r^2\sin(\theta)\cos(\theta)]|}{r}\le \frac{|r^2\sin(\theta)\cos(\theta)|}{r}\le r\to 0$$

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