Continuously Differentiable Real Function at Removable Singularity

Theorem
Let $f : \R \to \R$ be a real function.

Let $a \in \R$ be real number.

Let $f$ be continuous on $\R$ and continuously differentiable in $\R \setminus \set a$.

Suppose that $a$ is a removable discontinuity of $f'$.

That is, suppose the limit $\ds \lim_{x \mathop \to a} \map {f'} x$ exists.

Then $f$ is continuously differentiable at $a$.

Proof
By assumption, $\ds \lim_{x \mathop \to a} \map {f'} x = L$ exists.

By definition:


 * $\forall \epsilon \in \R_{>0} : \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map {f'} x - L} < \epsilon$

Let $h := x - a$.

Suppose $0 < h < \delta$.

Consider a closed interval $\closedint 0 h$.

By assumption, $f$ is continuous on $\closedint 0 h$ and continuously differentiable in $\openint 0 h$.

By Mean Value Theorem:


 * $\ds \exists \theta \in \openint 0 1 : \frac {\map f h - \map f 0} {h - 0} = \map {f'} {\theta h}$

Thus:


 * $0 < \size {\theta h} \le \size h < \delta$

and:

Hence:


 * $\ds \forall h \in \openint 0 \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$

Consider a closed interval $\closedint {-h} 0$.

By assumption, $f$ is continuous on $\closedint {-h} 0$ and continuously differentiable in $\openint {-h} 0$.

By Mean Value Theorem:


 * $\ds \exists \theta \in \openint 0 1 : \frac {\map f 0 - \map f {-h} } {0 - \paren {-h} } = \map {f'} {-\theta h}$

Thus:

$0 < \size {- \theta h} \le \size {-h} < \delta$

and:

Hence:


 * $\ds \forall y \in \openint {-\delta} 0 : \size {\frac {\map f y - \map f 0} {y - 0} - L} < \epsilon$

Altogether, it holds that:


 * $\ds \forall h \in \R : 0 < \size h < \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$

Hence, $f$ is differentiable at $a$:


 * $\ds \map {f'} a = L = \lim_{x \mathop \to a} \map {f'} x$.

By assumption, $f'$ is continuous on $\R \setminus \set a$.

Hence, $f$ is continuously differentiable on $\R$.