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Show that $f(x)= \begin{cases} x^2, \ x \in \mathbb{Q} \\ 0, \ x \not\in \mathbb{Q} \end{cases}$

has derivative only in $x=0$.

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2 Answers 2

up vote 3 down vote accepted

First show that the function is not even continuous at $x \neq 0$.

Consider $x \neq 0$.

If $x \in \mathbb{Q}$, consider the sequence $r_n = x - \dfrac{\sqrt{2}}n$. Show that even though $r_n \to x$, $$f(r_n) \to 0 \neq x^2 = f(x)$$

If $x \in \mathbb{R} \backslash \mathbb{Q}$, consider the sequence $q_n = \dfrac{\lfloor 10^n x \rfloor}{10^n}$. Show that even though $q_n \to x$, $$f(q_n) \to x^2 \neq 0 = f(x)$$

Hence, we can hope for a derivative to exist only at $x=0$.

At $x=0$, we have $$\dfrac{f(h) - f(0)}h = \begin{cases} h & \text{if }h \in \mathbb{Q}\\ 0 & \text{else}\end{cases}$$ and hence $$\left \vert \dfrac{f(h) - f(0)}h\right \vert \leq \vert h \vert$$ to conclude that the derivative exists and is $0$.

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Since $$\frac{f(x)}{x}= \begin{cases} x & \text{ if $x\in\mathbb{Q}$} \\ 0 & \text{ if $x\notin\mathbb{Q}$} \end{cases}$$ So $\lim_{x\to 0} f(x)/x=0$, and $f$ is differentiable at 0. And it is easily shown that $f$ is discontinous at for all $x\in\mathbb{R}-\{0\}$.

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Does $f$ has limit at any $x\in\mathbb R$?Thanks –  Babak S. Jan 18 '13 at 18:06
    
@Babak Sorouh $f$ has a limit only at $x=0$. –  tetori Jan 18 '13 at 18:10
    
+1 Thanks tetori. –  Babak S. Jan 18 '13 at 18:10

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