I found the following definition of the derivative interesting:
Let $f$ be a real function, $f$ is derivable at point $a$ if there is a $B \in \mathbb{R}$ s.t $$f(x)=f(a)+B(x-a)+R(x-a)$$ with $$R(x-a)\in o(x-a)$$ i.e., $$\lim_{x\to a}\frac{R(x-a)}{x-a}=0,$$ and we note $f'(a)=B.$
Now, here is my question: how to prove that $R$ is continuous, without assuming that $f$ is, nor using the standard definition of the derivative?
Thanks in advance, Tom
From the hypothesis,
$$f(x)=f(a)+B(x-a)+R(x-a)\implies R(0)=0$$
which holds for $x=a$ and
$$R(x-a)=o(x-a)\implies \lim_{x\to a}R(x-a)=0=R(0).$$
$$\lim_{x\to a } \frac{R(x-a)}{x-a}=0 \implies \lim_{x\to a} \frac{R(x-a)}{x-a}(x-a)=0\cdot 0=R(a-a)$$ as$$R(x-a)=\frac{R(x-a)}{x-a}(x-a)$$