# Showing that $\lim\limits_{x→0}\frac{x\sin(1/x)-\cos(1/x)}x$ does not exist without sequence

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.

• Why make things difficult for yourself like this? What is wrong with using sequences? – TonyK Nov 23 '18 at 15:24
• @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. – amir na Nov 23 '18 at 17:57

$$-1\leq \sin(1/x) \leq 1\implies -x\leq x\sin(1/x) \leq x$$ 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.
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.
• 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. – amir na Nov 23 '18 at 17:55