Given a function

$$ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & x < 0 \\ 0, & x=0 \\ 2 x^2 \sin\left(\frac{1}{x}\right), & x > 0 \end{cases} $$

how to compute its lower Dini derivative

$$Df(x) = \lim_{\varepsilon \rightarrow 0} \inf_{0 < \delta \le \varepsilon} \frac{f(x+\delta) - f(x)}{\delta}$$

using directly the definition? In particular, what is it at $x=0$? I can't wrap my head around it. There is a similar problem here, but that $2$ spoils everything.

  • $\begingroup$ You have $x^2 \sin\left(\frac{1}{x}\right)$ for $x \leq 0,$ but $x^2 \sin\left(\frac{1}{x}\right)$ is not defined for $x=0.$ You need to define $f(0)$ separately, and the value you use will affect the answer to your question. $\endgroup$ – Dave L. Renfro May 25 '18 at 15:12
  • $\begingroup$ Incidentally, if you define $f(0)=0,$ then $f'(0)$ exists and is equal to zero by using the definition of the derivative in the same way you would do if $2x^2$ were replaced by $x^2$ --- google "discontinuous derivative" + "squeeze". $\endgroup$ – Dave L. Renfro May 25 '18 at 15:46
  • $\begingroup$ @Dave L. Renfro My textbook says the result at $0$ should be $-2$! But I see you're right, $f$ is not even well-defined at $0$. $\endgroup$ – Rubi Shnol May 25 '18 at 16:08

(originally a comment, now extended and probably worth being an answer)

The lower right Dini derivate (what you want) would be $-2$ if the function was equal to $2x\sin(1/x)$ for $x>0,$ with $f(0)$ still equal to $0$ (and it doesn't matter what $f(x)$ is for $x < 0).$ See Calculating Dini derivatives for $f(x)=\begin{cases}x\,\sin{\left(\frac{1}{x}\right)} & x\neq 0\\ 0 & x=0\end{cases}$ for how the Dini derivates of the modified function can be found. Perhaps the place where you found this had incorrectly copied from the last page of this document or a similar one?

More generally, if

$$ g(x) = \begin{cases} ax \cdot \sin\left(\frac{1}{x}\right), & x < 0 \\ 0, & x=0 \\ bx \cdot \sin\left(\frac{1}{x}\right), & x > 0 \end{cases} $$

then the Dini derivates $D_{-},$ $D^{-},$ $D_{+},$ $D^{+}$ at $x=0$ are equal to $-|a|,$ $+|a|,$ $-|b|,$ $+|b|$, respectively.

And if $\alpha$ and $\beta$ are real numbers each greater than $1,$ and

$$ h(x) = \begin{cases} a|x|^{\alpha} \cdot \sin\left(\frac{1}{x}\right), & x < 0 \\ 0, & x=0 \\ bx^{\beta} \cdot \sin\left(\frac{1}{x}\right), & x > 0 \end{cases} $$

then the Dini derivates $D_{-},$ $D^{-},$ $D_{+},$ $D^{+}$ at $x=0$ are all equal to $0,$ and thus the two-sided derivative of $h(x)$ exists and equals $0$ at $x=0.$ (The reason for the absolute values when $x<0$ is to avoid problems with negative numbers raised to irrational powers.)


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