In my answer to this question (When we evaluate an indefinite integral of one variable, what area does this yield?) I wrote the following:

Now, if we divide both sides by $h$ we obtain the following: $$\lim_{h\to0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to0}\frac{f(x+h)+f(x)}{2}$$ but we can see that the expression on the left hand side is the definition of the derivative, $F'(x)$, for $F(x)$, our area accumulator function. So we can write $$F'(x)=\lim_{h\to0}\frac{f(x+h)+f(x)}{2}=\frac{f(x)+f(x)}{2}=\frac{2f(x)}{2}=f(x)$$

My question is, on the right hand side of the equality we basically say that $$\lim_{h\to0}f(x+h)=f(x)$$ However, why don't we say the same also on the left hand side; ie why don't we write (instead of $F'(x)$) $$\lim_{h\to0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to0}\frac{F(x)-F(x)}{h}=\lim_{h\to0}\frac{0}{h}=\infty$$ ie why doesn't the left hand side become meaningless if we are willing to write on the right hand side $\lim_{h\to0}f(x+h)=f(x)$?

Thank you for your help. If my derivation of the Fundamental Theorem of calculus is mistaken please tell me and help me correct it :)

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    $\begingroup$ What is your definition of a limit? $\endgroup$ – Physical Mathematics Sep 7 '20 at 20:31
  • $\begingroup$ Also $0/h = 0$ for $h \ne 0$ so $\lim_{h \to 0} 0/h = 0$. $\endgroup$ – Physical Mathematics Sep 7 '20 at 20:31
  • $\begingroup$ @KeeferRowan I am only in high school, but I would say that the way i currently understand a limit is what the expression approaches when a given variable is made to approach a given numerical value. Is that ok? $\endgroup$ – A-Level Student Sep 7 '20 at 20:32
  • $\begingroup$ @KeeferRowan even if it equals $0$ we still have a wrong equality, don't we? $\endgroup$ – A-Level Student Sep 7 '20 at 20:33
  • $\begingroup$ If you replace the numerator with its limit, you have to do the same for the denominator, and you obtain the indeterminate form $\frac 00$. $\endgroup$ – Bernard Sep 7 '20 at 20:43

As note din the comments by many, there are two things at issue here.

First, you cannot take a complex expression and only evaluate the limit to part of it, and then to the rest of it. If you have something like $$\lim_{x\to 0}\frac{x}{x},$$ the limit is $1$ (since the function takes the value $1$ at every $x\neq 0$); but you can't first evaluate the limit of the numerator (which is $0$), and then the limit of the resulting expression to claim that the limit is $0$.

So you cannot first do the limit of the numerator of $$\frac{F(x+h)-F(x)}{h}$$ and then do the expression. You need to do the whole expression. That is: you cannot compute the limit piecemeal within an expression: you must compute the limit of the whole, or of all its parts, at the same time.

Second, there is a "limit law" (actually a theorem for evaluating limits) that says:

Theorem. Let $g(x)$ and $h(x)$ be function, and assume that $\lim\limits_{x\to a}g(x)=L$ and $\lim\limits_{x\to a}h(x) = M$ both exist. Then:

  1. $\lim\limits_{x\to a}\bigl(g(x)+h(x)\bigr) = L+M$;
  2. $\lim\limits_{x\to a}g(x)h(x) = LM$;
  3. If $M\neq 0$, then $\lim\limits_{x\to a}\frac{g(x)}{h(x)} = \frac{L}{M}$.

This can be established fairly straightforwardly using the $\epsilon$-$\delta$ definition of a limit, though there is a bit of technical work to be done in 3 to make sure that $h(x)$ is bounded away from $0$. These limit laws are sometimes shorthanded as

The limit of a sum is the sum of the limits (if they both exist); the limit of the product is the product of the limits (if they both exist); the limit of a quotient is the quotient of the limits if the limit of the denominator is not $0$ (and both exist).

We can use these "limit laws" in the right hand side: the denominator is constant and has limit $2\neq 0$; so $$\lim{h\to 0}\frac{f(x+h)+f(x)}{2} = \frac{\lim\limits{h\to 0}\bigl( f(x+h)+f(x)\bigr)}{\lim_{h\to 0}2} = \frac{f(x)+f(x)}{2} = f(x),$$ using the limit law and the fact that $f(x)$ is assumed to be continuous everywhere in the interval, so $\lim\limits_{h\to 0}f(x+h) = f(x)$; in fact, we used limit law 1 to compute the limit of the numerator.

You cannot invoke this law/process in the left hand side, because there the denominator has limit $\lim\limits_{h\to 0}h = 0$. So you can't take the limits of the numerator and denominator separately. You have to deal with the entire expression in some way (obviously there are ways of doing it, or no derivative could ever be calculated; but you can't use the limit laws used on the right hand side).

(As an aside, note that while $\lim\limits_{h\to 0}f(x+h) = f(x)$ because we are assuming that $f$ is continuous, we have not proven that $F$ is continuous, so you also can't really justify the claim that $\lim\limits_{h\to 0}F(x+h)=F(x)$ unless you prove the continuity.)


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