Example · Undergraduate
Terms
| Term | This example | Why |
|---|
| differentiable |
✓ |
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is differentiable at every $x$: away from $0$, the product and chain rules give $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$, and at $0$, the difference quotient $\frac{h^2 \sin(1/h)}{h} = h \sin(1/h)$ tends to $0$ because $\abs{h \sin(1/h)} \leq \abs{h}$, so $f^{\prime}(0) = 0$.
|
| continuously differentiable |
✗ |
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is not continuously differentiable, because $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.
|
| smooth |
✗ |
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is not smooth, because $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.
|
| analytic |
✗ |
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is not analytic, because $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.
|
Distinctions using this example
- Differentiable vs. continuously differentiable: $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is the standard example of a differentiable function that is not continuously differentiable: the derivative exists at $0$, but the derivative $2x \sin(1/x) - \cos(1/x)$ oscillates between about $-1$ and $1$ near $0$, so the derivative is not continuous at $0$.
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