There seems to be a(t least one) serious (but rather common) misunderstanding.
When you say things like "the smallest value we can imagine," what do you mean? A natural answer (assuming that, by "smallest," you mean "closest to $0$") is $0$. However, context leads me to believe that you are referring to some nonzero value. Now, here's the problem: if $v$ is a non-zero value, then so is $\frac12v,$ and the latter is strictly smaller! Hence, there is no such thing as "the smallest nonzero value we can imagine."
Added: It seems you're also misunderstanding what a limit is. Again, this is a very common mistake to make! Within the last two weeks, I was explaining this to another user. Unfortunately, the question (and my answer to it) have since been deleted, so I can't link to it. Alas!
Let me first give you a few verbal definitions. After each verbal definition, I will give the common rigorous definition and connect the two.
We will be supposing hereinafter that $f$ is a function from $E$ to $\Bbb R$ for some subset $E$ of $\Bbb R.$
We will also be assuming that $x_0$ is a limit point of $E$ in $\Bbb R.$ Put into words, this means that $x_0$ is a point of $\Bbb R,$ and no matter how close we get to $x_0,$ we can always find a point of $E$ that is closer to (and yet distinct from) $x_0.$ Put rigorously: given any $\delta>0,$ there is some $x\in E$ such that $0<|x-x_0|<\delta.$ Here, $\delta$ tells us how "close" we need to get to $x_0$ and $|x-x_0|$ is the distance from $x$ to $x_0.$ So, $|x-x_0|<\delta$ says that $x$ is "even closer" than our required distance $\delta,$ while $0<|x-x_0|$ tells us that $x$ and $x_0$ are distinct. Note that $x_0$ may or may not be an element of $E.$ For example, $0$ is a limit point of the set of non-zero real numbers (of which it is not an element) and of the set of nonnegative real numbers (of which it is an element).
Now, we say that $L$ is a* limit of $f(x)$ as $x$ approaches $x_0$ if we can make sure that $f(x)$ is as close as we like to $L,$ simply by making sure that $x$ sufficiently close to (but not equal to) $x_0.$ Put rigorously: given any $\epsilon>0,$ there is some $\delta>0$ such that for any $x$ in $E$ with $0<|x-x_0|<\delta,$ we have that $|f(x)-L|<\epsilon.$ Here, $\epsilon$ tells us the desired "closeness" level between $f(x)$ and $L,$ so $|f(x)-L|<\epsilon$ says that $f(x)$ is as close to $L$ as we want it to be. Again, $0<|x-x_0|$ tells us that $x$ and $x_0$ are distinct,. Also, $\delta$ shows us the sufficient "closeness" level between $x$ and $x_0,$ so that $|x-x_0|<\delta$. shows that $x$ is sufficiently close to $x_0$ for our purposes. Depending on how close we want $f(x)$ and $L$ to be (that is, depending on how small $\epsilon$ is), we may need to change our requirement for how close $x$ and $x_0$ must be (that is, our $\delta$ may change).
*It turns out that, if there is a limit of $f(x)$ as $x$ approaches $x_0,$ then it is unique, and we denote it by $\lim\limits_{x\to x_0}f(x).$
So, what's the point? Well, let's consider what $\lim\limits_{x\to 0}x$ means, if anything.
The function $f(x):=x$ is readily defined on all of $\Bbb R$--that is, we have $E=\Bbb R$ in this case--and $x_0=0$ is readily a limit point of $\Bbb R.$ We can prove this by adapting the argument from my initial answer (before the "Added" part).
Now, for we want to know if there is some $L$ such that, given any $\epsilon>0,$ we can pick $\delta>0$ such that if $0<|x-x_0|<\delta,$ then $|f(x)-L|<\epsilon.$ Translating this into our particular situation, we want to know if there is some $L$ such that, given any $\epsilon>0,$ we can pick $\delta>0$ such that if $0<|x-0|<\delta,$ then $|x-L|<\epsilon.$ Looking at it that way, it should be clear that if we put $L=0$ and $\delta=\epsilon,$ then this works out just fine. The upshot is this: $$\lim\limits_{x\to 0}x=0.$$
This may come as a shock. After all, didn't we require $x$ to stay away from $0$ in our definition of limit? Well, yes, in part. We required our domain values to be distinct from $0$. Consequently, our range values were also required to be distinct from $0.$ However, the limit was not restricted! Not only does $0$ fit the definition of the desired limit, it is the only number that does!
Added: Let me go through a proof of limit uniqueness in a less formal way that will hopefully be easier to understand.
Claim: If $L$ is a limit of $f(x)$ as $x$ approaches $x_0,$ then it is the only number that is. Hence, $\lim\limits_{x\to x_0}f(x)$ is well-defined.
Heuristic Proof: Take $L'$ to be any number not equal to $L,$ so that the distance between $L$ and $L'$ is positive. Call this distance $d,$ and note that $\frac12d$ is positive. Hence, we have by definition of limit that, for $x$ sufficiently close (but not equal to) $x_0,$ $f(x)$ is within $\frac12d$ of $L.$ Consequently, for such $x,$ the distance from $f(x)$ to $L$ is less than half of the distance from $L'$ to $L.$ Hence, for such $x,$ we have that $f(x)$ is more than $\frac12d$ away from $L',$ and less than $\frac32d$ away from $L'$ (though the latter isn't important). In order to have $L'$ as a limit of $f(x)$ as $x$ approaches $x_0,$ though, we would need to be able to make $f(x)$ be less than $\frac12d$ away from $L'.$ Having showed this to be impossible, we find that for any $L'\ne L,$ $L'$ is not a limit of $f(x)$ as $x$ approaches $x_0.$
The approach above is basically a more general (and direct) version of the formalistic and particular proof I gave in the comments. There, I used proof by contradiction, and "$L$" there corresponds to "$L'$" here. From this claim, it follows that $\lim\limits_{x\to 0}x=0,$ in particular.
