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Question: Why does the derivative of $g(z) = \max\{0,z\}$ not exist at $z = 0$?

I know that for $z = 0$ we have $g'(z) = \lim\limits_{h\to 0}\dfrac{g(z+h) - g(z)}{h} = \lim\limits_{h\to 0}\dfrac{g(h)}{h} = \lim\limits_{h \to 0} \dfrac{h}{h} = 1,$ as $g(h) = h$ if $h \neq 0$. If the limit exists the function has a derivative, right?

Thanks in advance!

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    $\begingroup$ It might be helpful to graph $g(z)$. There's a corner at $z=0$. $\endgroup$ – carmichael561 Dec 29 '17 at 14:53
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    $\begingroup$ Your statement "$\lim\limits_{h\to 0}\frac{g(h)}{h}=\lim\limits_{h\to 0}\frac{h}{h}$" is false. It is only true if $h\to 0+$. $\endgroup$ – MPW Dec 29 '17 at 15:04
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For the derivatives to exist, you should check at both sides of $z=0$.

The right side derivative: $$\lim_{h \rightarrow 0+} \frac{g(h)}{h} = 1$$ (as you calculated)

But the left side derivative is not the same: $$\lim_{h \rightarrow 0-} \frac{g(h)}{h} = \lim_{h \rightarrow 0-} \frac{0}{h}=0$$ (Since $\max \{ 0, h \}$ when $h<0$ is $0$) Thus the derivative does not exist.

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