# An example of Gateaux derivative

In one of the Gateaux derivative examples mentioned here on page 5, it is mentioned that the function $f(x) = \lvert x\rvert$ has the Gateaux derivative defined at $x=0.$ In particular, it says that $$\lim_{\epsilon\to 0}\frac {\left\lvert {\epsilon h} \right\rvert} {h} = \left\lvert h \right\rvert$$

I feel that the above limit is not unique. It can be either $+h \text{ or} -h$. The Gateaux derivative is not defined at $x = 0$

• I would not trust your cited source. Dec 29, 2017 at 16:39

By definition $$d_hf(0)=\lim_{\epsilon\to0}\frac{f(\epsilon h)-f(0)}{\epsilon}=\lim_{\epsilon\to0}\frac{|\epsilon h|}{\epsilon}=\lim_{\epsilon\to0}\frac{|\epsilon| }{\epsilon}|h|.$$ You wrote $h$ in the denominater instead of $\epsilon$. And here you see that the limit doesn't exists because $$\lim_{\epsilon\to0^+}\frac{|\epsilon| }{\epsilon}|h|=\lim_{\epsilon\to0^+}|h|=|h|\neq-|h|=\lim_{\epsilon\to0^-}-|h|=\lim_{\epsilon\to0^-}\frac{|\epsilon| }{\epsilon}|h|.$$ So there is a mistake in your reference.