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Using the definition: $f’(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ i try to find the derivative of the absolute value function. My first two steps are, $f’(x)=\lim_{h\to 0} \frac{|x+h|-|x|}{h}$ rewritten as $\lim_{h\to 0} \frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}$ It’s at this point that I’m not sure about my last step, and if it’s correct then I’m lost on how to proceed in order to get the h cancellation to evaluate the limit. I apologize if this is a duplicate. Thank you.

EPDurfey
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    I'm afraid the easiest way (that I know of) is to distinguish cases $x\lt 0$, $x=0$, $x\gt 0$. –  Feb 04 '18 at 17:02
  • Thank you both, I will give that a look see. – EPDurfey Feb 04 '18 at 17:05
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    There is one answer in the dupe target which uses the $\sqrt{x^2}$ approach you’ve started. The trick is to rationalize the top (in general, you’ll want to rationalize differences of radicals, for these kinds of limit computations) – pjs36 Feb 04 '18 at 17:06
  • OT, but I just love this site. You all nailed it. – EPDurfey Feb 04 '18 at 17:15

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Although it is a duplicate, here is an alternate answer to the one given there:

\begin{eqnarray} \frac{d}{dx}|x|&=&\lim_{h\to0}\frac{|x+h|-|x|}{h}\\ &=&\lim_{h\to0}\frac{|x+h|-|x|}{h}\cdot\frac{|x+h|+|x|}{|x+h|+|x|}\\ &=&\lim_{h\to0}\frac{|x+h|^2-|x|^2}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{(x+h)^2-x^2}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{h(2x+h)}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{2x+h}{|x+h|+|x|}\\ &=&\frac{x}{|x|}=\frac{|x|}{x} \end{eqnarray}

John Wayland Bales
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