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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives: Example $\text E$