differential of $\| \cdot \|$ at $0$

Question

I know that the function : $x \mapsto \| x\|$ doesn't have a differential at $x = 0$.

Yet I don't see the problem with what I'm doing.

$f(0+h)-f(0) = \| 0+h\|- \| 0 \| = \|h \|$.

Here, I want to prove that : $\mathrm{d}f(0) : h \mapsto \|h\|$.

In order to proove that we may notice that :

$f(0+h)-f(0)-\|h\| = 0 = o(\|h\|)$, hence we have : $\mathrm{d}f(0) : h \mapsto \|h\|$.

What is the problem with what I've done ?

Answer 1

The derivative $df(0)$ is (if it exists) the unique linear map with $f(h)-f(0)-df(0)[h]=o(\Vert h \Vert)$. But $h\mapsto \Vert h \Vert$ is not linear.

Answer 2

$$\frac {f(0+h)-f(0)}{h} =\frac {\|h\|}{h}$$

$$ lim _{h\to 0} \frac {f(0+h)-f(0)}{h} = $$

$$ lim _{h\to 0} \frac {\|h\|}{h}$$ does not exist.

Thus $f$ is not differentiable at $x=0$