How do I disprove using $\epsilon - \delta $ definition of limits? I have only learned (in the morning), that how to use $\epsilon- \delta$ definition to "prove" a limit, thanks to the people here, that helped me learn how to pick my delta, and after 6+ hours of practising, I think I have gotten fairly good at that. 
I'd like to learn how to "disprove" using the same method. My book, which I'm self studying from, Bartle, doesn't have a lot on disproving. So I'd like some to help me out with this.
First of all, what I know: I understand to disprove, It'd be sufficient to pick a particular $\epsilon$, and show that for every $\delta >0$ there exists some $x$ belonging to the deleted delta nbd of $c$ (point at which I'm finding the limit) for which,  $|f(x)-L|≥ \epsilon$ this is more or less, negation of the definition.
So taking random example, let $f(x)= 2x+3$, disprove the $\lim\limits_{x \to 2} f(x)= 10$
So $|f(x)-10| = |2x-7|$
Now before taking random $\epsilon$ I'd like someone to point me out, how $\epsilon$ is chosen? Because clearly, the condition tells us, there exists "at least one" $\epsilon$, not "for all" $\epsilon$, so (I'm guessing) there are some specific $\epsilon$'s only out of which we have to pick one. 
Next up, how is that "particular" $x$ taken from the deleted nbd of $2$.
I did see some examples on the internet, but as I'm a beginner, given my limited understanding of real analysis, it becomes very hard to follow the arguments without explanation. Hence this question.
I know I'm asking a lot, so any help is appreciated
 A: Let me follow up with your "random" example, i.e. trying to disprove that $\lim_{x \to 2} f(x) = 10$ where $f(x)=2x+3$. As you say, $|f(x)-10| = |2x-7|$. 
So, we must find an appropriate value of $\epsilon > 0$, and using it we must prove that for every $\delta > 0$ there exists $x$ so that $0 < |x-2| < \delta$ and $|2x-7| \ge \epsilon$.
With existence proofs, there's often no general "formula" or "method". You just have to think, calculate, dig, hunt, rummage, use whatever you can come up with, and hopefully make an educated guess at an appropriate value of $\epsilon$. 
To hunt down an appropriate value of $\epsilon$, first I would use my intuition: when $x$ is close to $2$ then $|2x-7|$ is close to $3$. So, it seems appropriate that I might succeed if I choose $\epsilon$ to be some number less than $3$. That's still a lot of choices. But maybe I'll be lucky and the proof will be possible with any choice of $\epsilon < 3$.
With that in mind, I'll choose $\epsilon = 2$. So, for any $\delta>0$, I need to find a simultaneous solution to the two inequalities
$$0 < |x-2| < \delta \quad\text{and}\quad |2x-7| \ge 2
$$
The solution set of the second inequality is $(-\infty,2.5] \cup [4.5,+\infty)$ and I can see that this set contains numbers that are arbitrarily close to $2$, in particular it contains the entire interval $(2 - .5, 2 + .5)$. So if my $\delta$ is large then $x=2.4$ will probably do, whereas if my $\delta$ is small then $x=2 + \delta/2$ will probably do. So I'll try $x = 2 + \min\{.4,\delta/2\}$, and it works.
A: I think zooming out with a little generalization may help here.  Prove that a function cannot have two limits at a point.  

Proposition. If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} f(x) = M$, then $L=M$.

The idea is that $f(x)$ cannot be arbitrarily close to distinct numbers $L$ and $M$ as $x\to a$.  To get a contradiction, choose an $\epsilon$ such that $|f(x) - L| < \epsilon$ and $|f(x) - M| < \epsilon$ can't both be satisfied.
Now if you can show that $\lim_{x\to 2} (2x+3) = 7$, you know that $\lim_{x\to 2}(2x+3) \neq 10$.
To take the exercise a bit further, suppose you want to prove that function does not have a limit at all at a point.  This adds an extra quantified variable to the statement.  $\lim_{x\to a}f(x)$ does not exist if for all $L$, there exists $\epsilon>0$ such that for all $\delta > 0$, there exists $x$ such that $0 < |x-a| < \delta$ but $|f(x) - L| \geq \epsilon$.  Now show:

Proposition. If $\lim_{x\to a^{+}} f(x)$ and $\lim_{x\to a^-} f(x)$ exist and are distinct, then $\lim_{x\to a}f(x)$ does not exist.

A: Your example is the easy one because you already know how to prove $\lim_{x\to 2}(2x+3)=7$. To show that the limit is not $10$ choose an $\epsilon $ less than $|7-10|=3$. For example let $\epsilon=2$ and $$|f(x) - 10|=|2(x-2)-3|$$ and clearly if $|x-2|<1/2$ then we can't have $|f(x) - 10|<\epsilon=2$.
The more interesting problem is to show that $\lim_{x\to 0}\sin(1/x)$ does not exist. For this the common technique is not to use $\epsilon, \delta$ but rather choose sequences $a_n, b_n$ both tending to $0$ but $\sin(1/a_n),\sin(1/b_n)$ tending to different limits. 
