Is the definition of "limit of function" incomplete? On Wikipedia the definition of limit of a function $f$ such that it assigns an output $f(x)$ to every input $x$ is given as follows:

We say that the function has a limit $L$ at an input $p$, if $f(x)$ gets
closer and closer to $L$ as $x$ moves closer and closer to $p$.

But I have a problem with it; If $L$ (for concrete example say $5$) is chosen as limit of the function then can't $L-0.1$ ($4.9$) or $L-1$ ($4$) or $L+1$ ($6$) also be chosen as limit?
Let me explain what I mean.
If value of "input" is made to approach $p$ then, as given, the output will also approach $L$, and also $L-0.1$, $L-1$..... so what makes us choose only $L$ as the "limit"?
There is no special, explicit property, seems to be, mentioned which allows us to choose $L$ as the only "limit" and disregard other values(or does it?) like the fix difference between output, for given input, and the limit .
 A: When a function $f$ and a limit point $p$ are given, then the function limit itself $\lim\limits_{x \to p}f(x)$ cannot be chosen - the function limit, if it exists, is uniquely determined by the function itself and the point $p$ to which the variable tends.
Let me say, also, that all these words "tends", "gets closer", "closer and closer" are some kind of mathematical sleng - more good is to think about formal definition:
$$\forall \varepsilon >0, \exists \delta >0, \forall x, 0<|x-p|< \delta \Rightarrow |f(x)-L| < \varepsilon$$
so existence of limit is holding of inequality in appropriate conditions: "$x$ tends to $p$", "$x$ get closer to $p$" means first inequality and "$f(x)$ get closer to $L$", "$f(x)$ tends to $L$" means second.
A: The other answer gives the definition in formal mathematical symbols, but I would like to emphasize two things.
First, it is not necessary to understand those formal mathematical symbols at first, you can probably learn the meaning of limit more easily and effectively when the definition is written out in ordinary language, in this case in the English language, with just a few inequalities

*

*For every real number $\epsilon > 0$ there exists a real number $\delta > 0$ such that for every real number $x$, if $0 < |x-p| < \delta$ then $|f(x)-L| < \epsilon$.

Second, real understanding comes when you are asked to use this definition: either proving a particular limit formula; or applying some limit formula that is already known to be true, and using it to prove something else.
So let me go just a little bit further and explain generally how you can think of the process of proving a limit formula.
If $f(x)$, and $p$, and $L$ are given to you, and if you are asked to prove that $L$ is the limit of $f(x)$ as $x$ approaches $p$, here is what you have to actually do in order to prove it:

*

*Let a real number $\epsilon > 0$ be given.

*You must find an appropriate real number $\delta > 0$ and, using it, you must prove the implication: "if $0 < |x-a| < \delta$ then $|f(x)-L|<\epsilon$".

To really see this process in action on an intuitive level, I like to think of this as a challenge game.
Your adversary presents you with $\epsilon$, perhaps with an exact value like $\epsilon = .01$. And then the adversary says to you "I bet you can't make $|f(x)-L| < .01$, no matter how close $x$ is to $p$!!!"
So you take the challenge: using the given formula for $f(x)$, and the given value of $L$, and the actual numerical value $\epsilon = .01$ that the adversary gave you, you first figure out an actual numerical value of $\delta > 0$ (it might be as simple as solving an inequality). Then, using that value of $\delta$, you prove that if $0 < |x-a|<\delta$ then $|f(x)-L|<.01$.
You show this to the adversary who then says, with a pout, "Pshaw! That was too easy. I bet you can't do it with $\epsilon = .0001$!"
So you take up the challenge again: using the given formula for $f(x)$ and the given value of $L$, you figure out a value of $\delta > 0$, and you use it to prove that if $0 < |x-a|<\delta$ then $|f(x)-L|<.0001$.
Then your adversary says "You think you're so smart, here's the ultimate challenge. I bet you can't do it with $\epsilon = .0000000000000000000001$!"
By then you are losing your patience, and so you ignore this particular value of $\epsilon$, and instead you proceed like this: using just the symbol $\epsilon$ and the assumption that $\epsilon > 0$, you find a formula for $\delta > 0$, expressed in terms of $\epsilon$. Perhaps you find the formula using your previous experience with two particular numerical values of $\epsilon$; perhaps you find it by a more complicated mathematical process of solving the inequality $|f(x)-L| < \epsilon$. One way or another, using that formula for $\delta$, you then go on and use it to prove that if $0 < |x-a|<\delta$ then $|f(x)-L|<\epsilon$.
And your adversary disappears in a puff of smoke.

So, how might this whole adversarial process be described on an intuitive level? Perhaps something like this:

We can force $f(x)$ to be close to $L$ as we are asked to, by taking $x$ as close to $p$ as we like.

Or, in even fewer words, like this (i.e. like the Wikipedia quote):

$f(x)$ gets closer and closer to $L$, as $x$ gets closer and closer to $p$.

However, I hope that by now you can see that this is not a full and proper mathematical definition: without all of the quantifiers and absolute values and inequalities in their proper positions, this intuitive definition does not actually tell you what you have to do to prove or to apply limit formulas.
In order to acheive that, all of that "closer and closer" stuff has to be formalized with correct mathematical expressions using quantifiers, absolute values, and inequalities.
