When taking limits, how can we cancel the h in the denominator THEN assume it's 0? If we said it was zero from the beginning, we would be dividing by zero, which doesn't work

You don't set h to zero, but rather you let it approach zero. h, in this case, is as close to zero as possible without actually being zero. Thus, you can divide it out.

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This is a very good question, though, and is the motivation the "rigorization" of calculus in the 1800s.
As mentioned, we don't assume that h *is* 0. Instead, we're interested in how the expression behaves as we get close to 0 without actually attaining the value 0. Is there a specific number it approaches, for instance?

Now it's not immediately obvious wether that's the case while both numerator and denominator approach 0. But taking a limit has nothing to do with fractions and their denominator and numerator specifically. It just happens that in this case, we're interested in the limit of a fraction. But the fact that its numerator and denominator behave in a seemingly "bad" way shouldn't concern us that much. We only care about the value of the *entire* expression as a whole, and only while it's *close* to 0 without actually being 0. We don't ask: what value does the expression take at h=0? We ask: what value does the expression get close to while h isn't 0 yet? The trick is to simplify the expression in such a way that the latter question suddenly has the same answer as the former. This step is crucial: we do end up answering a question of the form "what value does the expression take?" But the original question is not of this form, and it only reduces to this form if the expression has been simplified! Many students forget what the original question was and just remember the simplified question, instead of the crucial step of turning the hard question into the simple one.
Just to provide an example, just graph y=x/x. It looks like y=1, except that there's a hole at x=0. But you know it's supposed to be 1 there as well. Seems overly simple, but it's no different from simplifying (x^(2)\-x)/(x-1) to x. Those are NOT exactly the same because the first one as a hole at x=1, and if you wrote they were equal, you should put (except when x=1) after it.

So

1) you're replacing the bad function with a continuous one that's the same except at the bad point

2) limit means specifically "yeah but just ignore the bad point"

So for the purposes of the limit, they are the same function. And the only actual limit law is that we know the limit for continuous functions.
at no point when taking the limit as h -> 0 of something do you ever set h = 0. that is the entire point of limits. what actually happens at h = 0 is completely irrelevant and has no effect on the value of the limit.
It is a good question!  For students just learning who don't mind if it seems a bit contradictory, you can say that you divide by h since you know it isn't zero, then you can eliminate the ones remaining because they are zero as a limit.
As already mentioned, we don’t take h=0 in this process ever. We take the limit as h->0.

However, have you come across the concept of continuity? It tells us that a function is continuous at a point x=a if and only if the limit of our function as x->a turns out to be the same as our function evaluated at x=a.

Thus we can use this fact and say that if your simplified expression 2x+h is continuous in h, then the limit is exactly the same as just substituting in h=0 in the end. This is why it looks like we set h=0.

As a counter-example, consider something like the continuous function f(x) = |x| = x*sign(x) where we have the discontinuous function sign(x) = -1,0, or 1 depending on whether x is negative, zero, or positive, respectively.

We see that f’(x) = limit as h->0 of [ (x+h) * sign(x+h) - x * sign(x) ] / h

We can simplify this to: f’(x) = limit as h->0 of [ x * (sign(x+h) - sign(x)) ] / h + sign(x+h)

Consider some cases , we treat h to be so small that x+h and x have the same sign …

* If x>0 then sign(x+h) = sign(x) = 1 so f’(x) = limit as h->0 of 1 which is a constant (so continuous in h) hence we can set h=0 and see that f’(x) = 1.

* If x<0 then sign(x+h) = sign(x) = -1 so f’(x) = limit as h->0 of -1 which is a constant (so continuous in h) hence we can set h=0 and see that f’(x) = -1.

* If x=0, then f’(x) = limit as h->0 of sign(h). Now this sign function is not continuous in h (precisely at h=0) so we cannot substitute h=0 into here to get the limit. In fact, because our limit depends on whether h is positive or negative, it means that it does not exist.

In this example we have shown that the continuous function f(x)=|x| is not differentiable at x=0, but it has a gradient of +1 for values x>0, and gradient -1 for values x<0.
you take a value of h that is extremely close to 0 so that it can be rounded down to 0, but not actually 0 so that you can divide by it

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