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! 
 A: Hint: Either $L=1$ or $L\neq1$. In the former case, when do we have $\sin(x)=-1$ for $x\in\Bbb R^+$? (Think setwise, not specific values). Now invert those numbers, and see that you have found infinitely many places in $(0,1)$ with $|\sin(1/x)-L|>1-\varepsilon$ for any $\varepsilon>0$. In the second case, instead solve for $\sin(x)=1$, and repeat a similar process.
A: Many ways and hints you've got are useful. This solution is good too:
Lemma1: The limit of a real function is unique if it exists.
Lemma2: The limit of a real function $f$ at $x=a$ is $L$ iff for any sequence $(a_n)$ which converges to $a$, sequence $f(a_n)$ converges to $L$.  
So if $\sin {1 \over x}$ has limit at $x=0$, for two sequences $(a_n)=({1\over n\pi})$ and $(b_n)=({2\over (4n+1)\pi})$ we must have:
$$\lim\limits_{n \to \infty} \sin {1 \over a_n}=\ \lim\limits_{n \to \infty} \sin {1 \over b_n}$$
Because both are the limit. Simple calculations show that LHS is $0$ and RHS is $1$, a contradiction.
A: Your approach is almost correct, but the order in which you introduce
your variables is confusing and may cause trouble.
You don't just find an $x$ and an $\varepsilon$ which may depend on $L$.
Of course $\varepsilon$ may depend on $L$, but $x$ can depend both
on $L$ and on $\delta$.
I find it sometimes helps to think of problems like this as an adversarial game,
like chess.
My opponent will make the first "move", choosing $L$. 
Then it is my "move": I must choose a value of $\varepsilon$.
After that, my opponent's "move" is to choose $\delta$.
I must then "move" by choosing an $x$ such that 
$0 < \left| x \right| < \delta  \wedge \left| {\sin {1 \over x} - L} \right| \ge \varepsilon $,
and then I win.
The pattern this game follows is that you read off the quantifiers from the
outside, working inward (left to right). Every universal $\forall$ is a
"move" by my opponent, and every existential $\exists$ is one of my "moves."
The problem is to show a perfect "strategy" that will win any game no matter
what "moves" my opponent makes.
Often it is helpful to give a strategy that handles some values of $L$ one
way and some values a different way.
In another answer you have a strategy that treats $L=1$ as one
case and $L\neq 1$ as a different case.
My first instinct looking at this problem was to come up with
a strategy for the case $L \geq 0$, and then look at the strategy for
the case $L < 0$. The two strategies are almost the same, as it turns out.
Edit: Here are more details.
What follows is not a detailed proof (I want to leave some of this
as an exercise!) but some very strong hints.
The intuition behind this comes from the graph of $\sin\frac1x$, which
oscillates between $1$ and $-1$ infinitely often as $x$ approaches $0$
from either side.
The condition $0 < |x| < \delta$ allows us to use positive or negative values
of $x$, but the positive values will be more than enough for our needs.
Note that 
$\sin u = 1$ for $u=\frac\pi2, \frac\pi2 + 2\pi, \frac\pi2 + 4\pi, \ldots$,
while $\sin u = -1$ for $u=\frac{3\pi}2, \frac{3\pi}2 + 2\pi, \ldots$.
Let $x = \frac1u$. If $u > \frac1\delta$, then $0 < x < \delta$.
So no matter how the opponent chooses $\delta$, as long as $\delta > 0$
there are plenty of ways to choose $x$ so that $\sin\frac1x = 1$ and
$0 < x < \delta$, or to choose $x$ so that $\sin\frac1x = -1$ and
$0 < x < \delta$.
So on the last "move", we'll always be able to choose
between $\sin\frac1x = 1$ and $\sin\frac1x = -1$, whichever we want.
If $L \geq 0$ and $\sin\frac1x = -1$, 
then $\left| \sin\frac1x - L \right| \geq \underline{\phantom{00}}$.
If $L < 0$ and $\sin\frac1x = 1$ , 
then $\left| \sin\frac1x - L \right| > \underline{\phantom{00}}$.
Fill in the blanks in those statements, and then
it should not be hard to think of an $\varepsilon$
that will work in each case, that is,
so that $\left| \sin\frac1x - L \right| > \varepsilon$.
As it turns out, $\varepsilon$ does not need to be a function of $L$
for this particular proof.
Having devised a strategy, all that remains is to prove, step by step,
that the strategy will "win" for any $L \geq 0$,
and also that it will "win" for any $L < 0$.
A: 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 
$$0<x<\delta\land\left|\sin\left(\frac1x\right)-L\right|\ge\epsilon.$$
A: 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 :)
A: Solution using epsilon-delta argument directly but like Game!:
My oponent chooses a limit for $\sin {1 \over x}$ at $x=0$ namely $L$, I choose $\varepsilon={1\over 2}$ then he chooses $\delta$.
But as you can see from my last answer (two sequences) there are two number $x_1$ and $x_2$ (there are infinitely of them but I need just two!) Such that
$$|{x_1}| < \delta \,\, , |{x_2}| < \delta \,\, ,\,\,\sin {1 \over x_1}=0\,\, , \,\,\sin {{1 \over x_2}=1}$$
$$\text {if}\, L\notin ({-1\over 2},{1\over 2})\,\text {I choose}\, x_1 \,\,\, \text{else I choose}\, x_2 $$
A: Here is my own solution inspired by adversarial game interpretation of Eq.(1) in question above which is explained in the answer by David K . Another answer by Yves Daoust was also useful for proper choice of $\varepsilon $ at the beginning of the game. Suppose that $L$ is chosen by the opponent, then at this stage its my turn to choose $\varepsilon $. What $\varepsilon $ shall I choose? Let me consider 
$$\varepsilon  = {1 \over 3}\tag{1}$$ 
In the next step, the opponent will choose a $\delta $. Now I may choose one $x$ such that
$$0 < \left| x \right| < \delta  \wedge \left| {\sin {1 \over x} - L} \right| \ge \varepsilon\tag{2}$$
holds and fortunately there are many such $x$'s. So my choice of $x$'s can be 
$$x = \left\{ \matrix{
  \text{any}\,\,x:\left| {\sin {1 \over x}} \right| \ge {1 \over 3} \wedge 0 < \left| x \right| < \delta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L = 0 \hfill \cr 
  \text{any}\,\,x:\left| {\sin {1 \over x} - L} \right| \ge {1 \over 3} \wedge \,0 < \left| x \right| < \delta \,\,\,\,\,\,\,\,\,\,\,\,L \ne 0 \hfill \cr}  \right.\tag{3}$$
To be more specific I can choose
$$x_n = \left\{ \matrix{
  {1 \over {{\pi  \over 2} + 2n\pi }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L = 0 \hfill \cr 
  {1 \over { - {{\left| L \right|} \over L}{\pi  \over 2} + 2n\pi }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L \ne 0 \hfill \cr}  \right.\tag{4}$$
By this choice of $x$, the first conditions, i.e., $\left| {\sin {1 \over x}} \right| \ge {1 \over 3}$ and $\left| {\sin {1 \over x} - L} \right| \ge {1 \over 3}$, will be satisfied for sure. It remains to satisfy $0 < \left| x_n \right| < \delta $. We can do this by choosing enough large $n$. To be more exact, we can prove
$$\forall \delta  > 0,\exists n:n \ge N \to \left| {{x_n}} \right| < \delta\tag{5} $$
Hence, no matter what the opponent choose I can ensure by my proper choices that $(2)$ will happen. So I will win anyway which means that the limit does not exist! :)
A: Take the sequence $x_n = \frac{2}{\pi n}$ and note that $x_n \to 0$ when $n\to \infty.$
Note however that $f(x_n) = 0$ if $n$ is even and $(-1)^n$ if $n$ is odd.
