Let $f: \Bbb{R} \to \Bbb{R}$ be defined as $f(x)=x^2 \; \forall x \in \Bbb{Q}$ and $f(x)=0 \; \forall x \in \Bbb{R} \setminus \Bbb{Q}$.

Question

Let $f: \Bbb{R} \to \Bbb{R}$ be defined as $f(x)=x^2 \; \forall x \in \Bbb{Q}$ and $f(x)=0 \; \forall x \in \Bbb{R} \setminus \Bbb{Q}$. Show that $f$ is differentiable at $x=0$ and also find $f '(0)$

My attempt:

We will consider $$\lim_{x \to 0}\frac{f(x) - f(0)}{x-0}$$

consider,

$|\frac{f(x) - f(0)}{x-0} - 0| \leq |\frac{x^2 - 0}{x-0} - 0| = |x| \; \; \forall x \in \mathbb{R}$

now,

$\forall \epsilon >0 $ taking $\delta= \epsilon$

if $0<|x|< \delta$ $\Rightarrow |\frac{f(x) - f(0)}{x-0} - 0| \leq |x| < \epsilon $

Thus $\lim_{x \to 0}\frac{f(x) - f(0)}{x-0}$ exists and equals $0$

Hence derivative exists at $0$ and is equal to $0$

Is my proof correct ?

Answer 1

Yes, it is correct. It was a good idea when you concluded that $|f(x)|\le x^2$ so that you didn’t need to analyse separately the rational and the irrational case.

Answer 2

The observation that for $x\ne 0$, $\left|\dfrac{f(x)}{x}\right|\leq\dfrac{|x|^{2}}{|x|}=|x|$ is helpful.