calculus - The non-existence of $lim limits_{x to 0} sin {1 over x}$

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!

Answer

Solution using epsilon-delta argument directly:
If $\sin {1 \over x}$ has limit at $x=0$ namely $L$, for $\varepsilon={1\over 2}$ we must have $\delta$ such that:

$$|{x - 0}| < \delta \Rightarrow |\sin {1 \over x} - L| < {1\over 2}$$
But as you can see from my last answer (two sequences) there are two number $x_1=\frac{1}{n\pi}$ and $x_2=\frac{2}{(4n+1)\pi}$ (there are infinitely of them but I need just two!) Such that
$$|{x_1}| < \delta \,\, , |{x_2}| < \delta \,\, ,\,\,\sin {1 \over x_1}=0\,\, , \,\,\sin {{1 \over x_2}=1}$$
So
$$|0 - L| < {1\over 2}\,\, ,\,\,|1 - L| < {1\over 2}$$

Or
$${-1\over 2} < L < {1\over 2}\,\, ,\,\,{1\over 2} < L < {3\over 2}$$
That's a contradiction if you accept :)