calculus - Finding the Derivative of |x| using the Limit Definition

Please Help me derive the derivative of the absolute value of x using the following limit definition.

$$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

I have no idea as to how to get started.Please Help.

Thank You

Answer

Since the absolute value is defined by cases,

$$|x|=\left\{\begin{array}{ll} x & \text{if }x\geq 0;\\ -x & \text{if }x\lt 0, \end{array}\right.$$
it makes sense to deal separately with the cases of $x\gt 0$, $x\lt 0$, and $x=0$.

For $x\gt0$, for $\Delta x$ sufficiently close to $0$ we will have $x+\Delta x\gt 0$. So
$f(x)= |x| = x$, and $f(x+\Delta x) = |x+\Delta x| = x+\Delta x$; plugging that into the limit, we have:

$$\lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x\to 0}\frac{|x+\Delta x|-|x|}{\Delta x} = \lim_{\Delta x\to 0}\frac{(x+\Delta x)-x}{\Delta x}.$$
You should be able to finish it now.

For $x\lt 0$, for $\Delta x$ sufficiently close to zero we will have $x+\Delta x\lt 0$; so $f(x) = -x$ and $f(x+\Delta x) = -(x+\Delta x)$. It should again be easy to finish it.

The tricky one is $x=0$. I suggest using one-sided limits. For the limit as $\Delta x\to 0^+$, $x+\Delta x = \Delta x\gt 0$; for $\Delta x \to 0^-$, $x+\Delta x = \Delta x\lt 0$; the (one-sided) limits should now be straightforward.