$\displaystyle\lim _{x\to \:a} f(x) =L $ well this means that :$f(x)$ can be made as close as we want to $L$ provided that $x$ is sufficiently close to $a$. Well this definitely means that: $$\forall \epsilon > 0,\ \exists \delta >0\ \text{ such that }\ |f(x)-L|<\epsilon,$$ provided that $0<|x-a|<\delta$.
But their is a question came to my mind:
Why have we chosen to define it that way? Why haven't we instead defined ''$f(x)$ can be made as close as we want to $L$'' means that: ''$|f(x)-L|=$ any number we want'' instead of ''$\epsilon -\delta$'' one? Why we've defined the limits in the way of neighborhoods not like in a way like what I've written? Because if we want to to make $f(x)$ close as we like to $L$, there is a must that the distance between them can be any number, which is $|f(x)-L|=$ any number we want.
There are really two subtleties in the definition.
First you need to get rid of the complications that arise due to "$=$" sign . As I said in the comment, you cannot make $|\sin(x)|=2$ for real numbers atleast. Does it mean that the limit does not exist ? . Ofcourse not, you must have learnt in high school calculus that $\sin(x)$ is a continuous function for all real numbers $x\in\Bbb{R}$. So notice that we can definitely make $|\sin(x)|<2$ . In fact we can make $|\sin(x)|<\text{any positive number}$ .
So you should note that your "any(positive) number" definition" has to do with the "epsilon" part of the definition.
Now arises the issue of locality. Typically you want the limit of the function at a point $x=a$ . (More rigorously in terms of set topology , at an accumulation point of the domain set of $f$ ) . So let us again return to the case of $\sin(x)$ and $a=0$ but let us restrict the domain to $[-\frac{\pi}{2},\frac{\pi}{2}]$ .
Now you would want to make sure that the values close to $0$ are being taken when you are near to $0$.
For example, if I choose $\epsilon=0.0001$ (Recall that $\epsilon$ has to do with the "any number" part).
Then I should not be angry if I see that $|\sin(x)|<0.0001$ does not hold if $x$ is near $\frac{\pi}{2}$ or $\frac{\pi}{4}$ . So for this $\epsilon$ , a choice of $\delta=0.8$ would be a bad one. However we can choose it to be such that $0<\delta<\sin^{-1}(0.0001)$ and see that it works.
That is we need to ensure that in a suitable "neighbourhood" of the point $0$ , we should have that the condition $|\sin(x)|<0.0001$ holds. Now obviously this "neighbourhood" would depend on our choice of $\epsilon$. Now as to why $|x-a|<\delta$ is chosen in the definition, the answer to it has to do with something called "open sets" and "basic open sets" in point set topology. If you read further into real analysis, you will get to the point set topology of the real line and you will learn that any open neighbourhood of a point $a$ can be obtained by taking union of sets of the form $\{x\in\Bbb{R}:|x-a|<\delta\}$ where $\delta$ is varied. Notice that the above set is what we call an open interval $(a,b)$. Here we are using $(a-\delta,a+\delta)$.
But topology aside, I think you can agree that these symmetric open intervals about a point $(a-\delta,a+\delta)$ really do capture the essence of what it means to be near "$a$" .
So $\epsilon$ takes care of the any positive number part. The inequality sign makes sure that we don't run into a stronger constraint of a equality which might not be possible to achieve. And $\delta$ makes sure that you are concerned with what's happening NEAR the point and not far away from it .