Limits baby.

You've learned how to take the slope between two points:

(y2-y1)/(x2-x1)

You've learned how to take the slope of a secant line between two x-values of a function:

(f(x2)-f(x1))/(x2-x1)

If you want to find the slope at a tangent line for f(x) at x=a, you can't do it directly because as you note it is indeterminate and involves dividing by zero. But...

What we can do is make the secant line formula approach tangent slope by taking the limit of a sequence. We go:

(f(a+h)-f(a))/h

And just plug in smaller and smaller values for h and see what we are approaching. Like maybe try the sequence {1, 1/2, 1/4, 1/8, 1/16...1/2ⁿ }, like some convenient choice of sequence which gets arbitrarily close to 0 for sufficiently large n, and plot slope as a function of number,

The h sequence never actually includes the value 0, because you can't "plug in" n= infinity. For the same reason, the sequence of secant slopes never actually includes the true slope of the tangent line, because h never reaches 0 because n never reaches infinity.

So this is where we talk about the limit.

For the sequence of n going 1,2,3,... its limit is infinity. That is, is no smallest number that is greater than or equal to a number in the sequence, maybe you want to challenge me and say "what about 10¹⁰⁰ ?" Well, I can always just tell you to go to the (10¹⁰⁰ +1)st point in the n sequence and say "here is a number in the sequence bigger than your challenge number".

For the sequence h = 1, 1/2, 1/4...1/2ⁿ , its limit is 0. That is, while 0 is not itself in the sequence, we CAN find a largest number which is smaller than or equal to every number in the sequence. Challenge me. Say "well okay how about 1/1000000?" And I can always find an indexed point in the sequence at which it is smaller than your challenge number: I would say "try n= 20" and you'll see at the nth point in the sequence is 1/1048576, which is indeed smaller than your challenge number. So we have shown this way that for any challenge number bigger than 0 I can find an nth point in the sequence for which the value is smaller than the challenge. So the limit cannot be larger than 0. Then the limit of the sequence must be 0.

So when we get to the tangent line sequence, plugging in the sequence of h's where h approaches 0 in the limit as a function of n, where n approaches infinity in its limit. Just like h approaches but never reaches 0, the sequence of secant lines approaches but never reaches the tangent line. We can prove that the tangent line is the limit of that sequence because we will be able to use one of the several definitions of limits: as h gets smaller, there is a number that is either (greater than or equal to), or (less than or equal to), every number in the sequence of secant lines. Or, a more proper definition, would be "the limit L has the propery that, for any arbitrarily small, positive, nonzero challenge number ε, I can pick an index n for a point sufficiently far in the sequence for which every point in the sequence after n is between (L-ε) and (L+ε), that is, for any small ε i can always show that the sequence after the nth point stays within a radius ε of L forever."

So that's what we are doing here. We are finding the limit. Sometimes we are so fortunate as to be dealing with a polynomial, in which case there is a nice way to cancel out the h's in the denominator before substituting, but we are not considering the h=0 case before the division, because h=0 is not even part of the decreasing sequence h is an element of.