1. ## differentiability of functions

f(x) = x^2 if xЭQ
f(x) = 0 otherwise.
We are studying differentiability, and my tutor said that this function was differentiable at 0 but not at any other point.

I can see how to show its differentiable at 0, but I dont know how I would go about proving that is isn't differentiable at any other point.

2. Originally Posted by nlews
f(x) = x^2 if xЭQ
f(x) = 0 otherwise.
We are studying differentiability, and my tutor said that this function was differentiable at 0 but not at any other point.

I can see how to show its differentiable at 0, but I dont know how I would go about proving that is isn't differentiable at any other point.

Take any point $\displaystyle x_0\neq 0$ , and choose now a sequence of rational points converging to it, and then a seq. of irrational points converging to it.

If the limit $\displaystyle \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ exists it must exist no matter how we choose to make $\displaystyle x\to x_0$ ...

Tonio

3. Im sorry but I don't really follow that.
If i take two sequences both converging to the same point, how does that relate to the second part? Sorry to be a pain.

4. The derivative, at x= a, of f(x) is defined as
$\displaystyle \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}$

Further, $\displaystyle \lim_{x\to a} g(x)= L$ if and only if $\displaystyle \lim_{n\to \infty} g(a_n)= L$ for every sequence $\displaystyle \{a_n\}$ converging to a.

Now, if there exist two sequences, $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$, converging to a, such that $\displaystyle \lim_{n\to\infty}\frac{f(a_n)- f(a)}{a_n-a}\ne \lim_{n\to\infty}\frac{f(b_n)- f(a)}{b_n- a}$ then the function is NOT differentiable at a. That is, if the two sequences have different limits, then the limit as h goes to 0 cannot exist and so the function is differentiable. For any a other than 0, take $\displaystyle \{a_n\}$ to be a sequence of rational numbers converging to a, and take $\displaystyle \{b_n\}$ to be a sequence of irrational numbers converging to a.

5. Originally Posted by HallsofIvy
The derivative, at x= a, of f(x) is defined as
$\displaystyle \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}$

Further, $\displaystyle \lim_{x\to a} g(x)= L$ if and only if $\displaystyle \lim_{n\to \infty} g(a_n)= L$ for every sequence $\displaystyle \{a_n\}$ converging to a.

Now, if there exist two sequences, $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$, converging to a, such that $\displaystyle \lim_{n\to\infty}\frac{f(a_n)- f(a)}{a_n-a}\ne \lim_{n\to\infty}\frac{f(b_n)- f(a)}{b_n- a}$ then the function is NOT differentiable at a. That is, if the two sequences have different limits, then the limit as h goes to 0 cannot exist and so the function is differentiable. For any a other than 0, take $\displaystyle \{a_n\}$ to be a sequence of rational numbers converging to a, and take $\displaystyle \{b_n\}$ to be a sequence of irrational numbers converging to a.
Why not using the same techniques show that $\displaystyle f$ is not continuous at any $\displaystyle x\ne 0$? This is easier in this case and the non-differentiability follows immediately.