Proof of $f(x):=|x|^p \text{ differentiable in } 0 \Leftrightarrow p>1$

Question

Exercise:

Let $p > 0, 0^p:=0$. Proove: $f(x):=|x|^p \text{ differentiable in } 0 \Leftrightarrow p>1$ .

I tried something that seems to be "to less".

$\Rightarrow$

$$f(x):=|x|^p \text{ differentiable in } 0 \Leftrightarrow \lim_{h \rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h} \in \mathbb{R} \text{ exists} \Leftrightarrow \lim_{h \rightarrow 0} \frac{|h|^p}{h} \in \mathbb{R} \text{ exists} \Rightarrow h^p > h \Rightarrow p>1$$

$\Leftarrow$

$$p > 1 \Rightarrow h^p > h \Rightarrow \lim_{h \rightarrow 0} \frac{|h|^p}{h} \in \mathbb{R} \text{ exists} \Rightarrow \text{f differentiable}$$

Would someone please have a look at it? Thanks in advance!

Answer 1

You're almost there. Note that $\left|\frac{|h|^p}{h}\right |=\left|\frac{|h|}{h}\cdot|h|^{p-1}\right|\le |h|^{p-1} \to 0$ as $h\to 0,$ as soon as $p>1.$