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Show that $$\lim_{x→0}\frac{x\sin(1/x)-\cos(1/x)}x$$ does not exist.

I understand that at $0$, the $\dfrac{\cos(1/x)}x$ term varies between $(-\infty , + \infty)$. But I want a complete formal proof that use the definition of limit ($ε, δ$) and not using sequences like ($2k\pi n$) or other techniques like l'Hopital, etc… How to write a formal proof for that?

Also I want to show that $\lim\limits_{x\to0} (x\sin(1/x) - \cos(1/x))$ does not exist without using any sequences and only using the definition of limit.

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  • $\begingroup$ Why make things difficult for yourself like this? What is wrong with using sequences? $\endgroup$ – TonyK Nov 23 '18 at 15:24
  • $\begingroup$ @TonyK I found that in a book that didn't say anything about sequences in the limit and derivative chapter. So I wanted to find a solution without sequence. $\endgroup$ – amir na Nov 23 '18 at 17:57
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$-1\leq \sin(1/x) \leq 1\implies -x\leq x\sin(1/x) \leq x$ and $-1\leq \cos(1/x) \leq 1$.

$-1-x\leq x\sin(1/x)-\cos(1/x) \leq 1+x \implies \displaystyle\dfrac{-1-x}{x}\leq \dfrac{x\sin(1/x)-\cos(1/x)}{x}\leq \dfrac{1+x}{x}$. We can show that as $x\rightarrow 0$ the limit is unbounded.

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Suppose it does exist. Then $ \lim_{x→0} x\sin(1/x)-\cos(1/x)=0$. Since $\lim_{x→0} x\sin(1/x)=0$, this would mean $\lim_{x→0}\cos(1/x)=0$ which is the same as: $\lim_{y→\infty }\cos(y)=0$. So the proof simplifies to showing this last limit is not zero.

One way of showing that $\lim_{y→\infty }\cos(y)$ does not exist without using sequences would be to use this result (good exercise to try to prove):

if $f:\mathbb{R}→\mathbb{R}$ is a periodic function and $\lim_{x→\infty} f(x)$ exists. Then $f$ is constant.

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  • $\begingroup$ How do you show the last limit is wrong, without any use of sequence. I can show for sequence $2\pi n$ it is 1 so it is obvious that the limit is not 0. But I cannot show that without sequence. $\endgroup$ – amir na Nov 23 '18 at 17:55
  • $\begingroup$ @amirna check my edit. $\endgroup$ – John11 Nov 23 '18 at 19:25

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