A function $f: \mathbb{R}\rightarrow \mathbb{R}$ is defined by $$ f(x)= \begin{cases} 0,& \text{if } ~x=0 ~~\text{or}~~ x ~\text{is irrational} \\ \frac{1}{q^3}, & \text{if}~~ x=\frac{p}{q} ~~ \text{where}~~ p\in \mathbb{Z}, q\in \mathbb{N}~~ \text{and} \gcd(p,q)=1. \end{cases} $$
Show that $f$ is differentiable at $0$ and $f'(0)=0$.
A hints to the problem is given as $0\le |\frac{f(x)}{x}|\le x^2$ for $x \ne 0.$ I don't know how to solve this problem. Please help.
When $h \in \Bbb R \setminus \Bbb Q$,
$\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=\lim_{h \to 0} \frac 0h=0.$
When $0 \neq h=\frac pq \in \Bbb Q$,
Consider $0 \le |\frac {f(h)}{h}| =|\frac { \frac 1{q^3}}{\frac pq}|=|\frac {1}{q^2 p}| \le |\frac {p^3}{q^2 p}|=|(\frac pq)^2|=h^2$.
What we have got is $0 \le |\frac {f(h)}{h}| \le h^2......(*)$
Hence using sandwich theorem in $(*)$, $\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=\lim_{h \to 0} \frac {f(h)}{h}=0.$
Since both of the above limits exist and are equal, $f$ is differentiable. Also $f'(0)=0$ as it can be seen.
