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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

T.D
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  • slightly related - https://math.stackexchange.com/questions/2321000/a-new-definition-of-derivative?noredirect=1&lq=1 – Calvin Khor Feb 20 '19 at 13:12

1 Answers1

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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$$R(x-a)=\frac{R(x-a)}{x-a}(x-a)$$

T.D
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  • Can you explain a bit more on the first one please ? / Pourriez-vous développer la première partie? – T.D Feb 20 '19 at 12:59
  • The left equality is true at $x=a$, by hypothesis. Substitute. This argument doesn't use the continuity of $f$. –  Feb 20 '19 at 12:59
  • Thanks, this is also good +1'ed – T.D Feb 20 '19 at 13:08