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 $\lim_{x \to a} r(x-a)=0$.