Show that $\sqrt{x^2+y^2}$ is not differentiable at the origin.

Question

This is very simple but I am having some trouble. Please indicate whether my proof is correct.

We want to show that $f(x, y) = \sqrt{x^2 + y^2}$. Is not differentiable at the origin. If it was differentiable, the partial derivatives at the origin would exist. However, for example,

$$ \lim_{h \to 0} \frac{f(h,0) - f(0, 0)}{h} = \frac{|h|}{h} = \begin{cases} 1, \quad \text{ if } h \to 0^+ \\ -1, \quad \text{ if } h \to 0^- \end{cases} $$ and hence the partial derivatives fail to exist.

Thanks in advance.

Answer 1

Yes, that is correct. Or you can say that the map$$\begin{array}{rccc}g\colon&\mathbb R&\longrightarrow&\mathbb R^2\\&x&\mapsto&(x,0)\end{array}$$is differentiable at $0$, that $g(0)=(0,0)$, but that $f\circ g$ is not differentiable at $0$ ($f\circ g$ is the absolute value function).