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Let $f:{\mathbb R} \to {\mathbb R}$ be defined by $f(x)= \begin{cases} x^2, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases} $ Show that f' is differentiable at $x=0$ and find f'(0).

Thus far I have,

If $x$ is rational: $\lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0.$

If $x$ is irrational, I have $\lim_{x \to 0} 0 = 0.$

How do I connect this to prove f' is differentiable at $x=0,$ and show f'(0)=0?

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    Techincally, you can't say "If $x$ is rational", given that $x$ is approaching zero. What you *can* do is consider $x$'s that approach $0$ only along the rationals, i.e., $\lim_{\stackrel{x\to 0}{x\in\mathbb{Q}}}\frac{f(x)-f(0)}{x-0}$but then you would have to argue that you can consider just this and $x$s approaching along irrationals.2012-04-02

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I don't think it's a good idea to split things up as you did. You could get the result from what you have, but it would take more work that using a more direct approach (or at least as much work).

You could try the following:

Set $ g(x)={f(x)-f(0)\over x-0}={f(x)\over x}. $ Then for $x\ne0$ $ g(x)=\cases{x,& $x$ rational,\cr 0,& $x$ irrational }. $

$f$ is differentiable at 0 if and only if $\lim\limits_{x\rightarrow0} g(x)$ exists. In this case, the value of the limit is f'(0). (You should suspect the value is 0 from what you've done.)

Can you prove this? Hint: note $ -|x|\le g(x)\le |x|; $ so try using the squeeze theorem.

Alternatively, use an (easy) $\epsilon$-$\delta$ argument.

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    @Quaternary Yes, that's correct.2012-04-02