Understanding linear approximation I am trying to understand the quoted bit below more or less rigorously.

$x \approx x_0 \implies f(x) \approx m(x - x_0) + f(x_0)$

I already tried once here below, but realized that method doesn't work:
Approximating a function at a value
Now I found a theorem (that I rewrote) that I think works.

Theorem:
A function $f: \mathbb R \to \mathbb R$ is differentiable at $x_0$ iff there's some number $m$ and a function $\varphi$ s.t.

*

*$\displaystyle{\lim_{h \to0}\varphi(h) = 0},$

*$\displaystyle{f(x_0 +h) = f(x_0) + mh + \varphi(h)h}$
Proof:
Suppose $f$ is differentiable. Then $f'(x_0) = \displaystyle{\frac{f(x_0 + h) - f(x_0)}{h}}$ exists. Consider $\displaystyle{\phi(h) = \frac{f(x_0 + h) - f(x_0)}{h} - f'(x_0)}$. This function satisfies the two conditions above.
Now assume there exist $m, \varphi$ s.t. the two conditions above are satisfied. Then $\displaystyle{f'(x_0) =\lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)}{h} = \lim_{h \to 0}\frac{mh + \varphi(h)h}{h} = \lim_{h \to 0}(m + \varphi(h)) = \lim_{h \to 0}m  + \lim_{h \to 0} \varphi(h) = m +0 = m.}$ Since $m$ is a real number that exists, so does $f'(x_0)$ meaning $f$ is differentiable $\blacksquare$
Deriving the original statement:
Replacing $h$ by $x - x_0$ in the second condition above we get $f(x) = f(x_0) + m(x - x_0) + \varphi(x - x_0)(x- x_0)$. If $x$ are extremely close to $x_0$, then $x - x_0$ is very small. Thus $\varphi(x - x_0)(x- x_0)$ is a tiny number meaning for $x$ close to $x_0$, we have $f(x) \approx f(x_0) + f'(x_0)(x - x_0).$
My questions:

*

*Is the proof above correct?

*Does the derivation make sense?


The statement is false. It becomes true if “function $\varphi$” gets replaced with “function $\varphi$ continuous at $0$ and such that $\varphi(0)=0$”. You actually used this in your proof when you wrote that $\lim_{h\to0}\varphi(h)=0$. So, your proof is correct if we see it as a proof of this new version of the theorem.
And, yes, the derivation is correct.
