How to prove $f$ is differentiable at $a$?

Question

Prove that if a function has a good linear approximation, then it is differentiable.

More specifically, let $a \in \mathbb R$. Let $f$ be a function that is continuous at $a$. Let $y = L(x)$ be an arbitrary line. Prove that $f$ must be differentiable at $a$. Assume that $$\lim_{x \to a} \dfrac{f(x) - L(x)}{x-a} = 0$$

Answer 1

I'll assume that $L(x)=\alpha x+\beta$, for some numbers $\alpha$ and $\beta$. Then $L(x)=\alpha(x-a)+\gamma$, where $\gamma=\alpha a+\beta$. We know that$$\lim_{x\to a}\frac{f(x)-\alpha(x-a)-\gamma}{x-a}=0.$$But this is the same thing as saying that$$\lim_{x\to a}\frac{f(x)-\gamma}{x-a}=\alpha.\tag{1}$$Since the limit $\lim_{x\to a}\frac{f(x)-\gamma}{x-a}$ exists and it is a real number and since $\lim_{x\to a}x-a=0$, $\lim_{x\to a}f(x)-\gamma=0$ too. But then $\gamma=\lim_{x\to a}f(x)=f(a)$ and therefore $(1)$ means that$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\alpha.$$In other words, $f$ is differentiable at $a$ (and $f'(a)=\alpha$).