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Just a quick question, this may or may not be a duplicate by the way. I've seen the proof of the trig functions not existing separately but I couldn't seem to find them multiplied together like in this problem. My question is could I just separate the two functions, prove that both limits do not exist and say the overall limit doesn't exist or is that the completely wrong way of approaching the problem?

EDIT: Here's my attempt with the use of $\frac{sin(\frac{2}{x})}{2}$ hint.

Proof:

Note that $\sin(\frac{1}{x})\cos(\frac{1}{x})$ = $\frac{\sin(\frac{2}{x})}{2}$. Let f(x) = $\frac{\sin(\frac{2}{x})}{2}$.

$\sin \theta = 1$ when $\theta = \frac{4k+1 \pi}{2}$ for $k \in \mathbb{Z}$

$\sin \theta = -1$ when $\theta = \frac{4k+1 \pi}{2}$ for $k \in \mathbb{Z}$

so $\sin(2/x_n) = 1$ when $x_n = \frac{4}{(4n+1)\pi}$ which equals $0$ when $n$ goes to infinity and $f(x_n) = 1$ for all $n$

$\sin(2/y_n) = 1$ when $y_n = \frac{4}{(4n-1)\pi}$ which equals $0$ when $n$ goes to infinity and $f(y_n) = -1$ for all $n$

It then follows that:

$\sin(2/x_n)/2 \implies f(x_n) = 1/2$

$\sin(2/y_n)/2 \implies f(y_n) = -1/2$

Since $\{x_n\}$ goes to zero $x_n \neq 0$ and $\{y_n\}$ goes to zero $y_n \neq 0$ but

$\{f(x_n)\} \rightarrow 1/2 \neq -1/2 \leftarrow\{f(y_n)\}$

$\lim_{x \to 0} f(x)$ does not exist by the sequential criterion.

Is this totally off?

Brevan Ellefsen
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Sky
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    It's equivalent to $\frac{\sin \frac{2}{x}}{2}$. Any help required now? – The Dead Legend Apr 13 '17 at 05:27
  • I apologize, I'm not sure what you're implying here. – Sky Apr 13 '17 at 05:30
  • Do you know how to prove that sin (1/x) discontinuous at x=o? – The Dead Legend Apr 13 '17 at 05:33
  • @Sky your updates look good, but seems awfully like overkill to me. – Brevan Ellefsen Apr 13 '17 at 06:49
  • @BrevanEllefsen What do you mean by overkill? – Sky Apr 13 '17 at 06:50
  • @Sky eh, I merely meant a few lines could be shortened and condensed a bit. Nevertheless, it looks good – Brevan Ellefsen Apr 13 '17 at 06:56
  • Btw, I helped clean up your MathJax :) Let me know if I misinterpreted your work anywhere. – Brevan Ellefsen Apr 13 '17 at 06:57
  • @BrevanEllefsen By the way, thank you for the edit. I was slowly getting through them a couple at a time. You did that so much faster than I did haha. And by overkill do you mean I went too detailed or did you mean my syntax was all over the place? – Sky Apr 13 '17 at 06:57
  • Sorry, I wrote that last comment before the page updated to show your comments. – Sky Apr 13 '17 at 07:02
  • @Sky Hmm... now that I look at it more, I would still say you could clean things up a lot by using the $\pm$ and $\mp$ operators to halve your lines, but some parts actually need a bit more justification in my eyes. I would say the part that needs the most justification is the lines "which equals $0$ as $n$ goes to infinity and $f(x_n)=1$ for all $n$". For all $n$ in what set? Be explicit here. Also, don't expect reader to understand what "sequential criterion" is in all cases, though from context I presume you mean that two (constant) sub-sequences have been found which differ in limiting... – Brevan Ellefsen Apr 13 '17 at 07:06
  • ... value so the limit of your original function must not exist. Oh, and no problem about the MathJax. I have a lot of experience with it by this point... I noticed you had a few lines where incorrect syntax was used, so I figured I could quickly help out :) – Brevan Ellefsen Apr 13 '17 at 07:06

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You could use the identity $ \sin(2/x)/2 = \sin(1/x)\cos(1/x) $ and then quote the result for $ \sin(1/x)$. As for separating the two functions and proving it that way, the answer is no. For example $ f(x) = 1 $ for $ x > 0 $ and $ f(x) = 0 $ for $ x \leq 0 $ and $ g(x) = 1 $ for $ x \leq 0 $ and $ g(x) = 0 $ for $ x > 0 $. Then neither of the limits exist as $ x\to 0 $ but the product $ f(x)g(x) = 0 $ does exist.

Infimum
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