Obviously the sequence $\left(\log (n)\sin(n\, \pi)\right)^{\infty}_{n\,\in\,\mathbb N}$ is $0$ for every $n$ but when I would like to take a limit of this sequence as $n$ approaches infinity I get stuck. On the one hand there is no reason why the value would be anything but $0$ but on the other hand it is $\infty \cdot 0$ which is indeterminate. How to solve this?

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    $\begingroup$ Do you know the $\epsilon - N$ formulation of sequence convergence? With that, it's unnecessary to work around the indeterminate form. $\endgroup$ – Nick Jan 22 '14 at 1:28
  • $\begingroup$ I know that there is something like: for every $\epsilon > 0$ there is a $N > 0$ such that if $n > N$ then $|a_n - 0|<\epsilon$ where $a_n$ is my sequence. But I'm quite clueless on how to apply this. Is it enough to state that for every $n$ the $|a_n |<\epsilon$ for every $\epsilon > 0$ since $a_n = 0$ for every $n$? $\endgroup$ – Vojtech Hauser Jan 22 '14 at 1:51
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    $\begingroup$ That is 100% correct. This is more convenient than trying to reason about the limit using knowledge of indeterminate forms, because it only requires you to know that for all natural numbers $n$ the values of your sequence are zero, which is easy to see. $\endgroup$ – Nick Jan 22 '14 at 2:07
  • $\begingroup$ I've never much used this definition but it quite makes sense not to try to fiddle with the indeterminate forms and use this instead, thanks. Pardon me for asking further but is there a way to work around the indeterminate form? I'm just curious. $\endgroup$ – Vojtech Hauser Jan 22 '14 at 9:49
  • $\begingroup$ In this case, I don't believe so. The problem isn't just that you have $\infty \cdot 0$ in the limit as $n$ goes to infinity--you also have to take the limit of $\sin$ as its argument goes to infinity, which is worse than indeterminate, it simply doesn't exist. $\endgroup$ – Nick Jan 22 '14 at 14:13

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