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So I want to prove that this converges to $0$:
$$\lim_{(x,y) -> (0,0)}\frac{\sin(xy) -xy}{xy\sqrt{x^2+y^2}}$$ We have that if $f(r\cos(\theta), r\sin(\theta)) = F(r)G(\theta)$ where $lim_{r->\infty}F(r) = 0$ and $G(\theta)$ is bounded, then the function converges to $0$.
$$f(r\cos(\theta), r\sin(\theta)) = \frac{\sin(r^2\cos(\theta)\sin(\theta)) -r^2\cos(\theta)\sin(\theta)}{r^3\cos(\theta)\sin(\theta)}$$ Now, my question is whether I can define: $$F(r) = \frac{1}{r}$$ $$G(\theta) = \frac{\sin(r^2\cos(\theta)\sin(\theta)) -r^2\cos(\theta)\sin(\theta)}{r^2\cos(\theta)\sin(\theta)}$$ I'm wondering since $G(\theta)$ has $r$ in there as well. Is that fine to treat it as a constant?

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$\sin(z)$ is an entire function fulfilling $$ \sin(z)=\sum_{n\geq 0}\frac{z^{2n+1}(-1)^n}{(2n+1)!} $$ hence in a neighbourhood of the origin $$ \frac{\sin z-z}{z}\sim -\frac{z^2}{6}. $$ It follows that in a neighbourhood of $(0,0)$ we have $$ \left|\frac{\sin(xy)-xy}{xy\sqrt{x^2+y^2}}\right|\leq C\left|\frac{(xy)^2}{\sqrt{x^2+y^2}}\right|\leq C\left|\frac{(xy)^2}{\sqrt{2|xy|}}\right|\leq D|xy|^{3/2}. $$

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use that $$\sqrt{x^2+y^2}\geq \sqrt{2|xy|}$$

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  • $\begingroup$ this is a way to evaluate the limit, but not an answer to OP's question as far as I can tell $\endgroup$ – Operatorerror Sep 15 '16 at 14:59
  • $\begingroup$ yes this is a way to compute the limit $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '16 at 15:00
  • $\begingroup$ downvoted because it does not address the question $\endgroup$ – Operatorerror Sep 15 '16 at 15:05
  • $\begingroup$ thx for the downvoting $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '16 at 15:06
  • $\begingroup$ it isn't clear to me you even read the post. edit: I removed it because I feel bad downvoting on this site $\endgroup$ – Operatorerror Sep 15 '16 at 15:06
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If you mean that you're going to take the derivative of $G$ or take the limit as $r\rightarrow 0$, then it's not OK to treat $G$ as a constant if it has $r$' in the expression.

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  • $\begingroup$ No, I actually meant to treat $r$ as a constant in $G(\theta)$ so I could take $G(\theta)$ as my bounded function which will end the proof. $\endgroup$ – Matam Sep 15 '16 at 17:16

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