In this proof, how do we know $R(0)=0$?

Question

This question concerns the best answer to another post on this website:How is the derivative truly, literally the "best linear approximation" near a point?.

The author writes this lemma and gives a proof.

Lemma: If a differentiable function $f$ can be written, for all $x$ in some neighborhood of $a$, as $$f(x) = A + B\cdot(x-a) + R(x-a)$$ where $A, B$ are constants and $R\in o(x-a)$, then $A=f(a)$ and $B=f'(a)$.

Proof: First notice that because $f$, $A$, and $B\cdot(x-a)$ are continuous at $x=a$, $R$ must be too. Then setting $x=a$ we immediately see that $f(a)=A$.

The proof then continues...

What I am confused about is the conclusion that $f(a)=A$. I do not see how the author concludes this. To have $f(a)=A$, we must have $R(a-a)=R(0)=0$. But how do we know $R(0)=0$?

We know that $R$ is continuous at $x=a$ and therefore at $0$. So we know that $\lim _{x \to 0}R(x)=R(0)$. But $R\in o(x-a)$, so $\lim_{x \to a} \frac{R(x)}{x-a}=0$. Letting $\epsilon=1$ and applying the definition of limit we see $-(x-a)<R(x)<(x-a)$, and by squeeze theorem $\lim_{x \to a} R(x)=0$, but that doesn't help because we need to find $R(0)$ and we haven't proven that $R$ is continuous at $a$ so we can't conclude $R(a)=0$. Is there some mistake here? Anyway, those are just my thoughts. Please provide feedback, and if you spot any error, please correct me.

The point of the post is this: how do we know $R(0)=0$?

peek-a-boo · Accepted Answer · 2022-06-17T00:20:55.030

0

Let $\rho(x)=R(x-a)$. Then $\rho\in o(x-a)$ is the correct hypothesis. This means by definition $\lim\limits_{x\to a}\frac{\rho(x)}{x-a}=0$. Also, $\rho$ is continuous at $a$, so \begin{align} R(0)&=\rho(a)=\lim_{x\to a}\rho(x)=\lim_{x\to a}\left(\frac{\rho(x)}{x-a}(x-a) \right)=0 \cdot 0=0. \end{align} It is the statement that $R\in o(x-a)$ which is an abuse of notation.

edited Jun 17 '22 at 00:20

answered Jun 17 '22 at 00:07

peek-a-boo

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But his claim is $R(x) \in o(x-a)$ as $x$ approaches $a$. He did not claim $R(x-a) \in o(x-a)$ as $x$ approaches $a$. – Jun 17 '22 at 00:13
@cfre yes, and I'm saying that's a mistake – peek-a-boo Jun 17 '22 at 00:13

score 0 · Answer 2 · answered Jun 17 '22 at 00:18

(1). Letting $x$ take values NOT equal to $a,$ we have $$f(a)=\lim_{x\to a}f(x)=$$ $$=\lim_{x\to a}A+B\cdot (x-a)+R(x-a)=$$ $$=\lim_{x\to a}A+B\cdot (x-a)+o(x-a)=A$$ because $f$ is continuous because $f$ is differentiable.

(2).We have $f(x)=A+B\cdot (x-a)+R(x-a)$ for every $x$ in a nbhd of $a,$ including the case $x=a,$ so $$f(a)=A+B\cdot (0)+R(0)=A+R(0).$$

In this proof, how do we know $R(0)=0$?

2 Answers2