Q:

If $\| f(x) \| \leq \| x \|^{\alpha}$, for $\alpha > 1$ then f is differentiable in $p=0$

Let $f: \mathbb{R}^k \rightarrow \mathbb{R}^m$, show that if $\| f(x) \| \leq \| x \|^{\alpha}$, for $\alpha > 1, \forall x \in \mathbb{R}^k$, then f is differentiable in $p=0$ with differential $df_0 = 0$.
I guess I'd simply have to show that:
$$ \lim_{h \rightarrow 0} \frac{\| f(x+h)-f(x) \|}{\|h\|} =\lim_{h \rightarrow 0} \frac{\| f(h)-f(0) \|}{\|h\|} = 0$$,
but I'm not sure how to do this and I don't know how the inequality is supposed to be of any help.  Any hints? 

A:

Yes, that is pretty much it.  The inequality gives $f(0) = 0$. So: $$0 \leq \frac{\|f(h)\|}{\|h\|} \leq \|h\|^{\alpha - 1} \implies 0 \leq \lim_{h \to 0}\frac{\|f(h)\|}{\|h\|} \leq 0 \implies \lim_{h \to 0}\frac{\|f(h)\|}{\|h\|} = 0, $$ because $\alpha - 1 > 0$.

