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For the absolute value function: $y = |x|$, how does the definition of limit apply:
enter image description here

The derivative of the function doesn't exist because the right hand derivative and the left hand derivative are not equal:

Right-hand derivative of $|x|$ at zero =
$$\lim_{x\to0^+} \frac{|0+h| - |0|}{h}$$
$$\lim_{x\to0^+} \frac{|h|}{h}$$
$$\lim_{x\to0^+} \frac{h}{h} = 1$$

Left-hand derivative of $|x|$ at zero =
$$\lim_{x\to0^+} \frac{|0+h| - |0|}{h}$$
$$\lim_{x\to0^+} \frac{|h|}{h}$$
$$\lim_{x\to0^-} \frac{-h}{h} = -1$$

My question is when h approaches $x$ from the negative part, why doesn't $h$ become negative in the denominator? If it is because, the denominator was formed by:

$$\lim_{x\to 0^+} \frac{f(x+h) - f(x)}{(x-h) - x}$$, and since $h$ approaches from the negetive side, it becomes positive, then, how would this justify the following function:
enter image description here
Where the function is defined only on the positive side.

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    $\begingroup$ All the $x\to0$ should be $h\to0$. And pease use \lim_{h\to0}. $\endgroup$ – Julián Aguirre Jan 26 '17 at 14:53
  • $\begingroup$ @JuliánAguirre, I didn't get you, kindly please explain me a bit. $\endgroup$ – bzal Jan 27 '17 at 12:34
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    $\begingroup$ The right hand side only has $h$'s, not $x$'s (except in the last one), and is $h$ the variable that goes to $0$; $x=0$ is fixed. $\endgroup$ – Julián Aguirre Jan 27 '17 at 14:14
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    $\begingroup$ $$ \begin{align} f'(x)&=\lim_{\color{red}{h\rightarrow\,0}}\frac{f(x+h)-f(x)}{(x+h)-(x)}=\lim_{\color{red}{h\rightarrow\,0}}\frac{f(x+h)-f(x)}{h} \\ &\color{white}{.} \\ f'(x_0)&=\lim_{\color{red}{x\rightarrow\,x_0}}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{\color{red}{h\rightarrow\,0}}\frac{f(x_0+h)-f(x_0)}{h} \end{align} $$ $\endgroup$ – Hazem Orabi Jan 29 '17 at 12:06
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"Why doesn't $h$ become negative in the denominator?" Answer: It does. I think you're making the common confusion that $-h$ is negative and $h$ is positive. But they're not. If $h<0$ then $-h$ is positive and $h$ is negative.

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