Finding values of $ \phi $ where $ f $ is differentiable at $ x = 0 $. If the function is defined as 
$$ 
f(x) =
\begin{cases}
|x|^\phi, & x \text{ is rational} \\
0, & x \text{ is irrational} 
\end{cases}
$$
where $ \phi > 0 $. For what values of $ \phi > 0 $ is the function $ f $ differentiable at $ x = 0 $? 
I suspect this occurs when $ \phi > 1 $ (is this correct?) and how does one prove this rigorously? 
 A: Well, note that if $f$ is to be differentiable at $x= 0$, then the limit $\lim_{h \to 0} \frac{f(h)}{h}$ is to exist, since the differential quotient at the point $x = 0$ is $\frac{f(0+h) - f(0)}h = \frac{f(h)}h$.
Now, the point is that if $\lim_{h \to 0} \frac{f(h)}{h}$ is to exist, then it better be equal to zero, since we may choose $h$ to approach $0$ via the irrationals, in which case the differential quotient is the constant zero sequence.

Make a very clever observation here : we have $0 \leq f(x) \leq |x|^{\phi}$ for all $x$. The first inequality is clear : the second follows, since both sides are equal if $x$ is rational, and if $x$ is irrational then the right hand side(of the second inequality) is zero.
Therefore, for any $h$, we have $0 \leq \frac{f(h)}{h} \leq \frac{|h|^{\phi}}{h}$.
If $\phi > 1$, then by the squeeze theorem, the limit of the central quantity $\frac{f(h)}{h}$ is equal to zero, since both sides of it go to $0$ as $h \to 0$. Consequently, $f$ is differentiable.
If $\phi  \leq 1$ then $\left|\frac{f(h)}{h}\right| = |h|^{\phi - 1}$ for $h$ rational. Therefore, approaching $h$  via the rationals for $\phi \leq 1$ gives a non-zero limit (or no limit) for the differential quotient at zero, hence $f$ is not differentiable. 
A: First off, it must be continuous therefore $\phi\ge 0$. Clearly $\phi=0$ is one answer. If $\phi >0$ and the function is differentiable in $0$ the following limit $$\lim_{h\to 0}\dfrac{f(h)-f(0)}{h}=\lim_{h\to 0}\dfrac{f(h)}{h}$$must exists and equals some real value $l$. If so, we have $$\lim_{h\to 0}\dfrac{f(h)}{h}=\lim_{h\to 0\\h\in\Bbb Q}\dfrac{f(h)}{h}=\lim_{h\to 0\\h\in\Bbb Q}\dfrac{|h|^\phi}{h}=\lim_{h\to 0\\h\in\Bbb Q}{|h|^{\phi-1}}$$also$$\lim_{h\to 0\\h\notin\Bbb Q}\dfrac{f(h)}{h}=0$$which means that $$\lim_{h\to 0\\h\in\Bbb Q}{|h|^{\phi-1}}=0$$and we must have $$\phi>1$$the the set of solutions is $$\{0\}\cup(1,+\infty)$$
A: Since clearly $$ \lim\limits_{x \to 0 ; x \not\in \mathbb{Q}} \frac{f(x)}{x} = 0$$
In order to have the differentiability, we must have (and it is enough) $$ \lim\limits_{x \to 0 ; x \in \mathbb{Q}} \frac{f(x)}{x} = 0$$
Which is clealy equivalent to $\phi > 1$
