A derivative of a real function is not always continuous
by Hidenori
Proposition
Show that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by
\[\begin{align*} f(x) &= \begin{cases} x^2\sin(1/x) & (x \ne 0), \\ 0 & (x = 0) \end{cases} \end{align*}\]is differentiable in $\mathbb{R}$, yet $f’$ is not even continuous at 0.
Solution
$f’(x) = 2x\sin(1/x) - \cos(1/x)$ when $x \ne 0$.
\[\begin{align*} \lim_{h \rightarrow 0} \frac{h^2 \sin(1/h) - 0}{h - 0} = 1. \end{align*}\]However, $\lim_{x \rightarrow 0} f’(x) \ne 1$, so $f’(x)$ is not continuous at $0$.
Subscribe via RSS