Right-Hand Derivative not Limit of Derivative from Right

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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:

\(\ds \map {f'_+} 0\) \(=\) \(\ds \lim_{h \mathop \to 0} \dfrac {h^2 \sin \dfrac 1 h - 0} h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} h \sin \dfrac 1 h\)
\(\ds \) \(=\) \(\ds 0\) Limit of $x \sin \dfrac 1 x$ at $0$

but:

\(\ds \map {f'} {0^+}\) \(=\) \(\ds \lim_{x \mathop \to 0^+} \paren {2 x \sin \dfrac 1 x - \cos \dfrac 1 x}\) which does not exist

$\blacksquare$


Also see


Sources