Why aren't all differential quantities equal?
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.