$\lim\limits_{x\rightarrow\infty} f(x)\ne L$ would mean that there is an $\epsilon>0$ such that for any $M>0$, there is an $x>M$ so that $|f(x)-L|\ge \epsilon$.
To use the above to show that $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist, you would have to show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$ for any number $L$.
For your purposes, with $f(x)=x\sin x$, let $L$ be any number. We will show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$. Towards this end, take $\epsilon=1$. Now fix a value of $M$. Using Alex's answer, you can find an $x>M$ so that $|f(x)-L|\ge1$.
Thus $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist.
(The limit might be infinite (it isn't, see Alex's answer again); but this is another matter...)
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.
Best Answer
Hint:
What is the negation of $$\forall \varepsilon>0\;\exists \delta>0\;\forall x\;\biggl(\lvert x\rvert<\delta\implies\biggl\lvert\frac1x-5\biggr\rvert<\varepsilon\biggr)?$$
Second hint:
Roughly said, the negation of an implication is a counter-example.