Right-Hand Derivative not Limit of Derivative from Right

Theorem
Let $f$ be a real function.

Let the right-hand derivative $f'_+$ of $f$ exist.

Then it is not necessarily the case that:
 * $\map {f'_+} a$

is the same thing as:
 * $\map {f'} {a^+}$

Proof
By definition:
 * $\map {f'_+} a := \ds \lim_{h \mathop \to 0} \dfrac {\map f {a + h} - \map f a} h$

while:
 * $\map {f'} {a^+} := \ds \lim_{x \mathop \to a^+} \map {f'} x$

Let:
 * $\map f x = \begin {cases} x^2 \sin \dfrac 1 x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$.

Then:

but: