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Could you please give a proof of the following statement?

If $f$ is differentiable then $f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ exists. This can alternatively be written $$f'(a) = \frac{f(x)-f(a)}{x-a} + r(x-a)$$ where the remainder function $r$ has the property $\lim_{x \to a} r(x-a)=0$.

Why is that the case? If the limit of $m(x)$ is a constant $z$, then obviously calculating the limit of $m(x) + n(x)$ will also be $z$ assuming $\lim n(x) = 0$. In this case:

$$f'(a) = \lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a} + r(x-a)\right)$$

however it doesn't explain why $f'(a)$ can be expressed as $ \frac{f(x)-f(a)}{x-a} + r(x-a)$ where $\lim_{x \to a} r(x-a)=0$.

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Let $h=x-a$. Then, we rewrite the given limit as $$\lim_{h\to0}\frac{f(a+h)-f(a)}h$$This is clearly just $f'(a)$! We define $$r(h)=f'(a)-\frac{f(a+h)-f(a)}h$$to express the difference between the derivative and this fraction (obviously defined for $h\neq0$). So, since$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h$$we have $$\lim\limits_{h\to0}r(h)=\lim\limits_{h\to0}\left(f'(a)-\frac{f(a+h)-f(a)}h\right)$$$$=f'(a)-\lim\limits_{h\to0}\frac{f(a+h)-f(a)}h$$$$=f'(a)-f'(a)=0$$

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  • $\begingroup$ And $r(h)$ is a different function for every point in the domain of $f$, right? The remainder, even though zero in the limit, can be bigger or smaller depending on which point we pick. $\endgroup$ Dec 20 '19 at 18:31
  • $\begingroup$ @user5539357 Precisely. $\endgroup$ Dec 20 '19 at 18:32
  • $\begingroup$ Ok, probably that's the source of my confusion, for some reason I assumed it is suggested that $r$ is the same for all points and I couldn't imagine how it might be true. $\endgroup$ Dec 20 '19 at 18:36
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$r(x-a) $ is not mysterious. We have $$r(x-a) =f'(a) - \frac{f(x) - f(a)}{x-a} $$ It is almost tautological that this goes to $0$ because that is the definition of the derivative.

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