# $\lim _{(x,y)\rightarrow (0,0)} \ln(\sin(x^2+y^2))=?$

I am working on this practice question for an upcoming exam.

I am not sure if I am oversimplifying this here (as the question is worth 3 marks):

In polar coordinates I have:

$\ln(\sin(r^2))$, so on any path, as $(x,y)\rightarrow(0,0)$, $r\rightarrow 0$.

For $r$ sufficiently close to 0, lim $r\rightarrow 0$ r$\sin(r^2) = 0$. Therefore, $\ln(\sin(r^2))\rightarrow -\infty$.

We are allowed to take $-\infty$ as a a valid limit.

Any help/hints would be greatly appreciated!

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Surely you do not mean to say that if $r$ is real close to $0$ then $\sin(r^2)$ is exactly $0$. – André Nicolas Jun 6 '12 at 0:32
Hi Andre, no I meant the limit of sin(r^2) = 0. Thanks. – JackReacher Jun 6 '12 at 0:39

If you want to establish the limit is $-\infty$, you can argue that for every $n\in\mathbb{N}$, you can find an $r\in(0,\infty)$ such that $\ln(\sin(r^2))<-n$.
Added: To be absolutely clear, this works here because $\ln(\sin(r^2))$ is monotonically increasing between 0 and a small positive number. You might be more comfortable arguing $\epsilon$-$\delta$ style too.
No, that establishes that the lim inf is $-\infty$. – Robert Israel Jun 6 '12 at 1:19
A little too stilted, imo. It's just a matter of adding $\,...\forall n\in\mathbb{N}\,\,\exists \delta_n>0\,\,s.t.\,\,r<\delta_n\longrightarrow \ln\sin(r^2)<-n\,\,$ – DonAntonio Jun 6 '12 at 1:38
You could just say that $\lim_{r \to 0+} \sin(r^2) = 0$ with $\sin(r^2) > 0$ for $\sqrt{\pi} > r > 0$, and $\lim_{s \to 0+} \ln(s) = -\infty$, so $\lim_{r \to 0+} \ln \sin(r^2) = -\infty$. – Robert Israel Jun 6 '12 at 2:00
... a more general statement being that if $\lim_{r \to a+} f(r) = b$ with $f(r) > b$ for $a < r < a + \epsilon$, and $\lim_{s \to b+} g(s) = c$, then $\lim_{r \to a+} g(f(r)) = c$ (and this includes the cases where $c$ is infinite). – Robert Israel Jun 6 '12 at 2:04