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I need to prove the following limit and I would like to have some feedback about my thought process as I'm still a beginner.

$$\lim_{(x,y)\to (0,0)} \frac{\sin^2(xy)}{x^2+y^2}=0$$

My proof:

I use polar coordinates and I get an indeterminate form $[\frac{0}{0}]$

$$ \lim_{\rho \to 0} \frac{\sin^2(\rho^2\cos(\theta)\sin(\theta))}{\rho^2} =\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho} \cdot \lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}$$

Solving

$$\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\leq\left|\frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\right|\leq\frac{\rho^2\lvert\cos(\theta)\rvert\lvert\sin(\theta)\rvert}{\rho}\leq \rho\to 0$$

regardless of $\theta$. Thank you for your help!

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    $\begingroup$ Note that as $x\to0$ we have that $\sin{(x)}=x+o(x)$ $\endgroup$ Commented Aug 8, 2019 at 13:31
  • $\begingroup$ Your answer and your thought process are correct. $\endgroup$
    – stochastic
    Commented Aug 8, 2019 at 13:31

3 Answers 3

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You may use the fact that $$ |\sin x|\le |x| $$ for all $x\in\mathbb R$.

Then, for $(x,y)\ne (0,0)$, $$ 0\le \frac{\sin^2(xy)}{x^2+y^2}=\frac{\sin^2(r^2\cos\vartheta\sin\vartheta)}{r^2}\le \frac{(r^2\cos\vartheta\sin\vartheta)^2}{r^2}=r^2\cos^2\vartheta\sin^2\vartheta\le r^2\to 0 $$ as $(x,y)\to (0,0)$.

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I would write $$\frac{\sin^2(xy)}{(xy)^2}\times\frac{(xy)^2}{x^2+y^2}$$ and now we use AM-GM inequality $$x^2+y^2\geq 2|xy|$$ so $$\frac{(xy)^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to Zero, if $x,y$ are tending to Zero.

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    $\begingroup$ Well doesn't this constitute an alternate approach? $\endgroup$
    – Toby Mak
    Commented Aug 8, 2019 at 13:38
  • $\begingroup$ No, i'm using my brain! $\endgroup$ Commented Aug 8, 2019 at 13:38
  • $\begingroup$ @Dr.SonnhardGraubner The question title clearly asks for an approach with polar coordinates. $\endgroup$ Commented Aug 8, 2019 at 13:39
  • $\begingroup$ But the title says also not ,do not use an alternate approach! $\endgroup$ Commented Aug 8, 2019 at 13:40
  • $\begingroup$ @Dr.SonnhardGraubner (+1) Fair enough. $\endgroup$ Commented Aug 8, 2019 at 13:42
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You are practically done. You have shown $$0\leq\left\vert\frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert\leq \rho$$ and thus, by the squeeze theorem $$\lim_{\rho\rightarrow 0} \left\vert\frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert =0$$ Since the absolute value is continuous, this means $$\left\vert\lim_{\rho\rightarrow 0} \frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert =0 \Rightarrow \lim_{\rho\rightarrow 0} \frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}=0$$ and you are done.

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