Prove that $f:\mathbb{R} \rightarrow \mathbb{R}$
$$f(x)=\begin{cases} 0, &x \notin \mathbb{Q} \\ x^2, & x \in \mathbb{Q} \end{cases}$$
has a derivative at $0$
The derivative exists if the limit
$$\lim_{x\rightarrow 0 } \frac{f(x)-f(0)}{x} \text{ exists} $$
For $x \in \mathbb{Q}$:
$$\lim_{x\rightarrow 0 } \frac{f(x)-f(0)}{x}= \lim_{x\rightarrow0}\frac{x^2-0}{x}=0$$
Now I have to find that this limit is $0$ if $x \notin\mathbb Q$ right?
How do I prove that?
With $\epsilon, \delta:$
$x \in \mathbb{R}$.
We show that $f'(0)$ exists and $f'(0)=0$.
Since $f(0)=0$, we have for real $x$:
$|\dfrac{f(x)-f(0)}{x}| \le |\dfrac{x^2}{x}|=|x|.$
Let $\epsilon >0$ given.
Choose $\delta = \epsilon$ , then
$|x| \lt \delta$ implies
$|\dfrac{f(x)-f(0)}{x}-0| \le |x| \lt \epsilon,$.