Differentiable function/D in R/Linear approximation/Fact/Proof

Proof

If ${}f$ is differentiable, then we set

${}s:=f'(a)\,.$

Then the only possibility to fulfill the conditions for ${}r$ is

${}r(x)={\begin{cases}{\frac {f(x)-f(a)}{x-a}}-s{\text{ for }}x\neq a\,,\\0{\text{ for }}x=a\,.\end{cases}}\,$

Because of differentiability, the limit

${}\operatorname {lim} _{x\rightarrow a,\,x\in D\setminus \{a\}}r(x)=\operatorname {lim} _{x\rightarrow a,\,x\in D\setminus \{a\}}{\left({\frac {f(x)-f(a)}{x-a}}-s\right)}\,$

exists, and its value is ${}0$. This means that ${}r$ is continuous in ${}a$.
If ${}s$ and ${}r$ exist with the described properties, then for ${}x\neq a$ the relation

${}{\frac {f(x)-f(a)}{x-a}}=s+r(x)\,$

holds. Since ${}r$ is continuous in ${}a$, the limit on the left-hand side, for ${}x\rightarrow a$, exists.

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