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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 $0$.2012-06-06
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    Hi Andre, no I meant the limit of sin(r^2) = 0. Thanks.2012-06-06

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You intuition is correct but we need to get your justification a little more solid.

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.

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    No, that establishes that the lim inf is $-\infty$.2012-06-06
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    ...which is sufficient when your function is monotonic in a neighborhood of the point you are limiting to.2012-06-06
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    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\,\,$2012-06-06
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    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 $\lim_{r \to 0+} \ln \sin(r^2) = -\infty$.2012-06-06
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    ... 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).2012-06-06