Since we know that in a good linear approximation, L ( x ) = f...

One definition of having a good linear approximation for $f$ at $a$ is
$f(x)=f(a)+c(x-a)+ϵ(x-a)$
where $ϵ(x-a)$, the error term, is a function that satisfies the following
$\underset{x\to a}{lim}\frac{ϵ(x-a)}{x-a}=0$
Then, that $f$ is differentiable basically follows immediately, by subtracting $f(a)$, dividing by x−a and taking the limit as $x-a$ showing that the function is differentiable at $a$, with value $c$.
$$\begin{gathered} f(x)=f(a)+c(x-a)+ϵ(x-a)⟹f(x)-f(a)=c(x-a)+ϵ(x-a) \\ ⟹\frac{f(x)-f(a)}{x-a}=c+\frac{ϵ(x-a)}{x-a} \end{gathered}$$
and thus taking the limit
$$\begin{gathered} \underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=c+\underset{x\to a}{lim}\frac{ϵ(x-a)}{x-a} \\ ⟹f^{\prime }(a)=c \end{gathered}$$