TITLE: The non-existence of $\lim \limits_{x \to 0} \sin {1 \over x}$
QUESTION [5 upvotes]: Regarding to the definition of the limit of a function, employing a step by step approach using tautologies in logic, here it was proved that the limit of a function $f(x)$ does not exist at a point $x=a$  if and only if
$$\forall L,\exists \varepsilon  > 0:\left( \forall \delta  > 0,\exists x:(0 < \left| x - a \right| < \delta  \wedge |f(x) - L| \ge \varepsilon \right))\tag{1}$$
or
$$\nexists L:\left( {\forall \varepsilon  > 0,\exists \delta  > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta  \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)\tag{2}$$
Statements $(1)$ and $(2)$ are logically equivalent. I want to use just $(1)$ to prove that $\lim\limits_{x \to 0} \sin {1 \over x}$ does not exist. However, proofs using $(2)$ can also be interesting!
My thought
I just know that I must find an $\varepsilon $ which may depend on $L$ and one $x$ which may depend on both $L$ and ${\delta}$ such that for every $L$ and $\delta  > 0$ we must have
$$0 < |x| < \delta  \wedge \left| \sin {1 \over x} - L \right| \ge \varepsilon \tag{3} $$
I don't know how to proceed!

REPLY [3 votes]: It should be clear to you that for any $\epsilon<1$ and $L$ you can choose a $\theta$ such that
$$\left|\sin(\theta)-L\right|\ge\epsilon.$$ (Indeed, one of $-\frac\pi2$ and $\frac\pi2$ will do the trick.)
Then you can choose $x=\dfrac1{\theta+2k\pi}$ as small as you want by increasing the integer $k$ (take $k>\frac{1-\theta\delta}{2\pi\delta}$), ensuring 
$$0<x<\delta\land\left|\sin\left(\frac1x\right)-L\right|\ge\epsilon.$$