Proof
If
is
differentiable,
then we set
-
![{\displaystyle {}s:=f'(a)\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc00d8b5b20659975549cee94fbfdcc470a6accd)
Then the only possibility to fulfill the conditions for
is
-
![{\displaystyle {}r(x)={\begin{cases}{\frac {f(x)-f(a)}{x-a}}-s{\text{ for }}x\neq a\,,\\0{\text{ for }}x=a\,.\end{cases}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29c733d57bca0a56229faffe058079b4a61d5798)
Because of differentiability, the limit
-
![{\displaystyle {}\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)}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbad8d01355a24efabb238b123638f4aa461ef97)
exists, and its value is
. This means that
is continuous in
.
If
and
exist with the described properties, then for
the relation
-
![{\displaystyle {}{\frac {f(x)-f(a)}{x-a}}=s+r(x)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78d8f87bbf0aab85f010ad29e1f1cb40dd53f5b1)
holds. Since
is continuous in
, the limit on the left-hand side, for
, exists.