For what $ \alpha \in \mathbb R$ is $ |x|^\alpha $ differentiable in $x=0$?

Question

I came across the following question:

For what $ \alpha \in \mathbb R$ is $ |x|^\alpha $ differentiable in $x=0$?

What I have tried:

Since for $ \alpha = 1 $ is clearly non-differentiable in $x=0$, let's consider $ \alpha < 1 $ and $ \alpha > 1 $.

If $ \alpha < 1 $:

\begin{align} \lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} &= \lim_{h\to 0^+} \frac{|0+h|^\alpha-|0|^\alpha}{h} \\ &= \lim_{h\to 0^+} h^{\alpha-1} \\ &= +\infty \end{align}

So there is no need to check the case $\lim_{h\to 0^-}$.

After doing the same for $ \alpha > 1 $ we get $\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = 0$. And from the left side:

\begin{align} \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h} &= \lim_{h\to 0^-} \frac{|0+h|^\alpha-|0|^\alpha}{h} \\ &= -\lim_{h\to 0^-} \frac{(-h)^\alpha-0}{-h} \mbox{ (piecewise definition of abs)} \\ &= -\lim_{h\to 0^-} (-h)^{\alpha-1} \\ &= 0 \end{align}

So I concluded that $ |x|^\alpha $ is differentiable in $x=0$ for $ \alpha > 1 $ and non-differentiable for $ \alpha ≤ 1 $.

Yet after thinking a bit more two questions came into my mind:

a) Isn't it a problem that if $ \alpha $ is irrational $x^\alpha$ is defined only for $x>0$?

b) What about the $0^0$ case?

Answer 1

Your conclusions are correct.

a) You're looking at $|x|^\alpha$; since $|x|>0$, rational powers of negative numbers are not an issue.

b) The case $\alpha=0$ can be dealt with separately easily.