Modulus of x is not Differentiable Proof

In this section, we will prove that the absolute value of x is not differentiable at the point x=0. In other words, the function |x| is not differentiable at x=0.

Absolute Value of x is not Differentiable at 0

The function f(x)=|x| is defined as follows:

$|x|=\begin{cases} x, & \text{ if } x\geq 0 \\ -x, & \text{ if } x< 0 \end{cases}$

Question: The function $f(x)=|x|$ is continuous at $x=0$ but not differentiable at $x=0$.

Answer:

First we will discuss the continuity of f(x)=|x|.

We have

$\lim\limits_{x \to 0-} f(x)=\lim\limits_{x \to 0-} (-x)=0$

Again,

$\lim\limits_{x \to 0+} f(x)=\lim\limits_{x \to 0+} x=0$ and $f(0)=0$

Thus, we obtain that

$\lim\limits_{x \to 0-} f(x)=\lim\limits_{x \to 0+} f(x)=f(0)$

$\therefore$ The function f(x)=|x| is continuous at x=0.

Next, we will discuss the differentiability of f(x)=|x|.

By the definition of the derivative, we have

$f'(x)=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$

$\therefore f'(0)=\lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}$

$=\lim\limits_{h \to 0} \frac{f(h)}{h} \quad [\because f(0)=0]$

Now, the left-hand derivative of  f(x) at x=0 is

$Lf'(0)=\lim\limits_{h \to 0-} \frac{f(h)}{h}$

$=\lim\limits_{h \to 0-} \frac{-h}{h}$

$=\lim\limits_{h \to 0-} (-1)$ $=-1$

On the other hand, the right-hand derivative of  f(x) at x=0 is

$Rf'(0)=\lim\limits_{h \to 0+} \frac{f(h)}{h}$

$=\lim\limits_{h \to 0+} \frac{h}{h}$

$=\lim\limits_{h \to 0+} (1)$ $=1$

As the left-hand derivative $\neq$ the right-hand derivative, we conclude that f(x)=x is not differentiable at x=0.

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