# How to prove differentiability?

1. Dec 12, 2007

### Reid

1. The problem statement, all variables and given/known data
Given an interval I. The function f goes from I to the real line. Define f as f(x)=x^2 if x rational or 0 if x is irrational. show that f i differentiablle at x=0 and find its derivative at this point, i.e. x=0.

2. Relevant equations
I have a given lead on this. That |f(x)/x|<|x|. But I'm not sure how this should help me. I have started with the definition and I have tried to determine a certain epsilon for which this is true. But I don't know how.

3. The attempt at a solution
I have shown that the function is continuous at this point and therefore know that the derivative might exist. But when it comes to the actual differentiability I have that (according to the definition),
given a positive epsilon there exists a delta. For x satisfying the inequality
0<|x-c|<delta, the inequality |(f(x)-f(c))/(x-c)-L|<epsilon holds. L then is called f's derivative in the point c.

I'm not sure if this counts as an attempt. But this problem is really bugging me.

2. Dec 12, 2007

### Office_Shredder

Staff Emeritus
the derivative is $$lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} = lim_{x \rightarrow 0} \frac{f(x)}{x}$$ as f(0)=0 by definition. But notice f(x)=x2 or 0, so f(x)/x is either 0 or x. See what you can do with that

3. Dec 12, 2007

### VietDao29

Okay, so your function f is defined like as follow:

$$f(x) : = \left\{ \begin{array}{ll} x ^ 2 \quad & , \mbox{if } x \in \mathbb{Q} \\ 0 \quad & , \mbox{if } x \notin \mathbb{Q}\end{array} \right.$$

So that means, for all x:
$$0 \leq |f(x)| \leq x ^ 2$$

Divide all by |x| > 0, (since |x| > 0, the signs won't change) we have:

$$0 \leq \left| \frac{f(x)}{x} \right| \leq \left| \frac{x ^ 2}{x} \right|$$

$$\Rightarrow 0 \leq \left| \frac{f(x)}{x} \right| \leq | x |$$ (1) (this is true for x <> 0)

Now, the derivative at x = 0, of the above function is:

$$f'(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0}$$

$$= \lim_{x \rightarrow 0} \frac{f(x) - 0}{x - 0}$$ (0 is rational, so f(0) = 0)

$$= \lim_{x \rightarrow 0} \frac{f(x)}{x}$$

Now, using the inequality (1) above can you show that the limit exists? And what is that limit? Note that:

$$|a| \rightarrow 0 \Leftrightarrow a \rightarrow 0$$, i.e a tends to 0, if and only if |a| tends to 0.

So, can you go from here? :)

Last edited: Dec 12, 2007
4. Dec 13, 2007

### Reid

Yes, I know where to go from there.

You are amazing! Thank you! :)