For $\alpha > 1$, show that $f:\mathbb{R}^n \to \mathbb{R}^n$, given by $$f(x)=\rVert x\rVert^\alpha$$ is differentiable at the origin.
What I did:
What is necessary is to find a linear transformation such that $\lim_{h\to 0}\frac{f(h)-f(0)-Th}{\rVert h \rVert}=0$. $f(h)-f(0) = \rVert h\rVert ^\alpha$. So, because $\alpha>1$, I guess that T=0 so that $\lim_{h\to 0}\frac{f(h)-f(0)-Th}{\rVert h \rVert}=\lim_{h\to 0}\frac{\rVert h\rVert ^\alpha}{\rVert h \rVert}=\lim_{h\to 0}\rVert h\rVert^{\alpha-1}=0$.
Is this correct?