Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \in \mathbb{Q^c}$. Show that f is differentiable at $x=0$, and find $f'(0)$
My Attempt:
let $c \in \mathbb{Q}$. Then $\lim_{x \to c}\frac{x^2 -c^2}{x-c}$ = $\lim_{x \to c}{x+c} = 0 + 0 =0$. I don't know if the idea of doing this through this is right. In particular, now as I type it, I realise that it should be 2c instead of 0 since its not the value of function we're talking about.
Can anyone please guide as to how I should approach this proof?
Thank you.
The value of $c$ that you are interested is $0$.
$f(x)$ doesn't take the value $x^2$ all the time.
$$\lim_{x \to 0} \left|\frac{f(x)-f(0)}{x-0}\right|=\lim_{x \to 0} \left|\frac{f(x)}{x}\right| \le \lim_{x \to 0}\left|\frac{x^2}{x}\right|=\lim_{x \to 0}|x|=0,$$
hence
$$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=0$$