If $f'(0)$ does not necessarily exists, can $f'_+(0)$ be defined as $\lim\limits_{n\to +\infty}n\left[f\left(\frac{1}{n}\right)-f(0)\right]$?

Question

Some book says that:

If $f'(0)$ does not necessarily exists, then $f'_+(0)=\lim\limits_{n\to \infty}n\left[f\left(\dfrac{1}{n}\right)-f(0)\right](n \in \mathbb{N})$ does not hold necessarily.

I'm confused about that. Is it true?

Answer 1

Yes. Let $$ f(x) = \begin{cases} 1/q, &x = p/q \quad [p\in \mathbb Z^\times, q\in \Bbb N^*, \gcd(p,q)=1],\\ 0, & x \in \mathbb R \setminus \mathbb Q \bigcup\{0\}, \end{cases} $$ Then clearly $$ \frac {f(1/n)- f(0)} {1/n} = 1, $$ but actually $f$ is not right-handed differentiable at $0$ thus not differentiable at $0$, since when $ x >0$, $$ \frac {f(x) - f(0)}x = \begin{cases} 1, & x \in \mathbb Q^+\\ 0, & x \in \mathbb R^+ \setminus \mathbb Q, \end{cases} $$ and the limit $$ \lim_{x \to 0^+} \frac {f(x)- f(0)}x $$ does not exist.

Answer 2

My View

Thanks for your enlightment. I have an idea for the problem, now.

Consider the function$$f(x)=\begin{cases}x\sin\dfrac{2\pi}{x},& x\neq 0;\\0,&x =0. \end{cases}$$

We may verify that $f(x)$ is continous at $x=0$, but it has not derivative even single-side derivative there. Since $$\frac{f(0+\Delta x)-f(0)}{\Delta x}=\sin\frac{2\pi}{\Delta x},\tag 1$$when$\Delta x \to \pm 0$, $(1)$ tends to no limit. But if we take $\Delta x=\dfrac{1}{n},n \in \mathbb{N},$ we have$$\frac{f(0+\Delta x)-f(0)}{\Delta x}=\sin 2n\pi\equiv 0,\tag 2$$when $n \to \infty$,$(2)$ has a limit $0$.

Obviously,$(1)$ and $(2)$ contradicts each other. We want to ask the reason why the contradition comes out. In fact, according to Heine theorem, to change the variable $x \to a$ into some sequence $x_n \to a$ in the limit expression is conditional. The conditon is just that $\lim\limits_{x \to a}$ exists. If $\lim\limits_{x \to a}$ does not exist, you can't change it into $\lim\limits_{x_n \to a}$, which is the reason.