Prove $\lim_{x\to0}2x\sin(\frac{1}{x})-\cos(\frac{1}{x})$ does not exist at $x=0$. Prove $\lim_{x\to0}2x\sin(\frac{1}{x})-\cos(\frac{1}{x})$ does not exist at $x=0$.
Does this need an $\epsilon-\delta$ proof, or is knowing it oscillates at 0 enough?
 A: can you use the Heine throem?
let $\dfrac{1}{a_n}=2k\pi$ and $\dfrac{1}{b_n}=2k\pi+\dfrac{\pi}{2}$ then the limit has 2 different results 
so the limit does not exist.
If a $\epsilon-\delta$ proof is necessary i would write afterwards.
let $\epsilon=1$    and $f(x)=2x\sin(\dfrac1x)-\cos(\dfrac1x)$
for any $\delta>0$ there exist a $x'$ and $x''$ $\in$ $U^o(0,\delta)$ and $\dfrac1{x'}=([\dfrac{1}{\pi\delta}]+1)\pi$  and $\dfrac1{x''}=([\dfrac{1}{\pi\delta}]+\dfrac32)\pi$
s.t. $丨f(x')-f(x'')丨=丨2x'\sin(\dfrac1{x'})-\cos(\dfrac1{x'})-2x''\sin(\dfrac1{x''})+\cos(\dfrac1{x''})丨\ge丨2x'\sin(\dfrac1{x'})--2x''\sin(\dfrac1{x''})丨+丨\cos(\dfrac1{x''})-\cos(\dfrac1{x'})丨\ge 丨\cos(\dfrac1{x''})-\cos(\dfrac1{x'})丨>\epsilon=1$
A: We see that $$\lim_{x \to 0} 2x \sin \frac {1}{x}=0 $$(well-known result and can be proved by Squeeze theorem.).
Now I will show that $$\lim_{x \to 0} \cos \frac {1}{x}$$ does not exist. 
For $\epsilon-\delta$ definition,use the following procedure. 
Choose $\epsilon =1$ . Let $f(x)= \cos \frac {1}{x}$.
Take $\frac {1}{x_1}=2n \pi$ and $\frac {1}{x_2}=(2n+1) \pi$ where $n \in \Bbb N.$ We see that $|f(x_1)-f(x_2)|=2 >1=\epsilon$ whereas $|x_1-x_2|<\delta$ which can be easily done by taking sufficiently large $n.$ 
