Prove that a function is not differentiable with $\epsilon - \delta$

Question

We have this function

$$f(x) = \begin{cases} x^2, & \text{if $x$ $\in \mathbb{Q}$} \\[2ex] 0, & \text{if $x$ $\in\mathbb{I}$} \end{cases}$$

We have to determine where the function is differentiable and calculate the derivative. Then we need to determine where is not differentiable and prove this with $\epsilon - \delta$

Attempt:

We define: $$\text{$h(x)=0$ and $g(x)=x²$, then $h(0)=0=g(0)=f(0)$}$$ Now we have: $$\text{$h(x)$ ≤ $f(x)$ ≤ $g(x)$, for all x$\in\mathbb{R}$}$$ $$\text{that is equal to: $h(x)-h(0)$ ≤$ f(x) - f(0) $≤$ g(x) - g(0)$}$$ $$\text{And without losing generality assuming that $x>0$:}$$ $$\text{$\frac{h(x)-h(0)}{x}$ ≤ $\frac{f(x)-f(0)}{x}$ ≤ $\frac{g(x)-g(0)}{x}$}$$ $$\text{Finally: $\lim_{x\to 0} \frac{h(x)-h(0)}{x}$ ≤ $\lim_{x\to 0} \frac{f(x)-f(0)}{x}$ ≤ $\lim_{x\to 0} \frac{g(x)-g(0)}{x}$}$$ $$\text{By the squeeze theorem we have that:}$$ $$\lim_{x\to 0} \frac{f(x)-f(0)}{x}=0$$ $$\text{So $f$ is differentiable at $c=0$ and $f'(0)=0$}$$

The problem comes when we have to show that is not differentiable at any other point. I managed to do it with limits but I don't know how to put in into $\epsilon - \delta$

$$\text{Let $x\in\mathbb{Q}$ and $x≠0$. Then $f(x)=x²$. Now we know that exist a sequence $(y_n)_{n\in\mathbb{N}}$ of irrational numbers such that:}$$ $$\lim_{n\to \infty}y_n = x$$ $$\text{Also we have that $f(y_n)=0$ for all $n$, because $y_n$ is irrational, but:}$$ $$\lim_{n\to \infty}f(y_n) = f(x) = x²$$ $$\text{So we can see that:}$$ $$\lim_{n\to \infty}f(y_n) ≠ f(y_n)$$ $$\text{This implies that $f$ is not continuous at $\mathbb{I}$, therefore $f$ is not differentiable at $\mathbb{I}$}$$ $$\text{The same thing works to prove that $f$ is not continuous at $\mathbb{Q}$ so $f$ is not differentiable at any point, except $x=0$.}$$

As I said the problem comes when I have to write that second thing with $\text{$\epsilon - \delta$ }$, cause I don't really know how to start. Someone has any ideas? Thanks to everyone.

Answer 1

Allow $(x_n)$ be a sequence of rational numbers such that $x_n \to \sqrt{2}$.

Fix $\epsilon > 0$. First notice that $|x_n - \sqrt{2}| < \delta \implies |x_n + \sqrt{2}| = |x_n - \sqrt{2} + 2\sqrt{2}| \leq |x_n - \sqrt{2}| + 2\sqrt{2} < \delta + 2\sqrt{2}$.

So $|x_n - \sqrt{2}||x_n + \sqrt{2}| < \delta(\delta + 2\sqrt{2})$.

Allowing $\delta < \frac{\epsilon}{1 + 2\sqrt{2}}$ for small enough $\epsilon$ (i.e. if $\epsilon < 1 + 2\sqrt{2}$) produces

$|f(x_n) - 2| = |x_n-\sqrt{2}||x_n + \sqrt{2}| < \delta(\delta + 2\sqrt{2}) < \delta(1+2\sqrt{2}) < \frac{\epsilon}{1+2\sqrt{2}}(1+2\sqrt{2}) = \epsilon$

Thus, $\lim_{x_n \to \sqrt{2}} f(x_n) = 2$, however $f(\sqrt{2}) = 0$.

It is not continuous, thus not differentiable.

Answer 2

If a function is differentiable, it is continuous. Since $f$ is discontinuous for $x \neq 0$ it cannot be differentiable for $x \neq 0$.

If $x \neq 0$ and $x$ is irrational you can find rationals $q_n \to x$ and then $f(q_n) = q_n^2$ but $f(x) = 0$. Then ${f(q_n)-f(x) \over q_n - x } = {q_n^2 \over q_n-x}$.

If $x \neq 0$ and $x$ is rational you can find irrationals $\alpha_n \to x$ and then $f(\alpha_n) = 0$ but $f(x) = x^2 \neq 0$. Then ${f(\alpha_n)-f(x) \over \alpha_n - x } = {x^2 \over \alpha_n-x}$.

In particular, the ratio is not bounded as $n \to \infty$, so it is straightforward to use an $\epsilon$-$\delta$ argument to show that it is not differentiable at $x$ (if it was differentiable, the ratio would be bounded).