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I am currently studying with notes that our professor gave us and there we have to check if a function is continous: Now let $f(x,y) = 0$ for $(x,y) = (0,0)$, $f(x,y) = \sqrt{|xy|}\sin\left(\frac{1}{x^2+y^2}\right)$ for $(x,y) \neq(0,0)$. We learned that you could transform $x$ and $y$ into polar coordinates and if $\lim_{r\to 0} \sqrt{|r^2|}\sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin\left(\frac{1}{(r(cos(\varphi))^2+(r(\sin(\varphi))^2}\right)$ = 0 and then it differentiable in from all sides and thus continuous.

Now my professor wrote that $\lim_{r\to 0} r\sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin(\frac{1}{r^2})$ = 0 since $\lim_{r\to 0} \sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin(\frac{1}{r^2})$ is bounded but this does not make any sense in my eyes since $\sin\left(\frac{1}{r^2}\right)$ is not defined and thus can't be bounded... Where did I make my mistake?

Thank you

user
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James
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2 Answers2

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For every real number $x$, $\sin(x)\in[-1,1]$. Therefore$$(\forall r\in\mathbb{R}\setminus\{0\}):\sin\left(\frac1{r^2}\right)\in[-1,1]$$and so, yes, your professor is right.

José Carlos Santos
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  • But $\lim_{x\to 0} sin(1/x^2) = \lim_{x\to \infty} sin(x^2) $ and this isn't defined for the same reason that $\lim_{x\to 0} sin(1/x) = lim_{x\to \infty} sin(x)$ is not defined? – James Aug 10 '18 at 14:11
  • @James That doesn't change the fact that you wrote that “since $\sin\left(\frac1{r^2}\right)$ is not defined and thus can't be bounded…” My answer proves that that claim is wrong. Anyway, since both $\sqrt{|\cos(\varphi)\sin(\varphi)|}$ and $\sin\left(\frac1{r^2}\right)$ are bounded, it is indeed true that$$\lim_{r\to0}r\sqrt{|\cos(\varphi)\sin(\varphi)|}\sin\left(\frac1{r^2}\right)=0.$$ – José Carlos Santos Aug 10 '18 at 14:12
  • Ok and even if its not defined it is still in the Intervall [-1,1]. Thank you – James Aug 10 '18 at 14:15
  • @James I'm glad I could help. – José Carlos Santos Aug 10 '18 at 14:16
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Recall that $\forall \theta \in \mathbb{R}$ we have that $|\sin \theta| \le 1$ and therefore

$$0\le \left|\sqrt{|xy|} \sin\left(\frac{1}{x^2+y^2}\right)\right|\le \sqrt{|xy|}\to 0$$

user
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