# Modulus of x-a is continuous at x=a but not differentiable

This page will discuss the continuity and differentiability of the absolute value of $x-a$, that is, of the function $|x-a|$, at the point $x=a$.

Note that the function $|x-a|$ is defined as follows:

$|x-a| = x-a$ if $x\geq a$

$=-(x-a)$ if $x<a$

## Continuity of |x-a| at x=a

Let us now discuss the continuity of the modulus function $|x-a|$ at $x=a$.

Write $f(x)=|x-a|$

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

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

and $f(a)=0$

So we get that

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

Thus the function f(x)=|x-a| is continuous at x=a.

## Differentiability of |x-a| at x=a

We will now show the modulus of $x-a$ is not differentiable at $x=a$.

Let $f(x)=|x-a|.$

Note that the function $|x-a|$ will be differentiable at $x=a$ if the following limit $L$ exists.

$L=\lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h}.$

The left-hand limit $=\lim\limits_{h \to 0-} \dfrac{f(a+h)-f(a)}{h}$

$=\lim\limits_{h \to 0-} \dfrac{-(a+h-a)-0}{h}$

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

The right-hand limit $=\lim\limits_{h \to 0+} \dfrac{f(a+h)-f(a)}{h}$

$=\lim\limits_{h \to 0+} \dfrac{(a+h-a)-0}{h}$

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

As both the left-hand and the right-hand limit exist and are not equal, we conclude that the limit L does not exist. Thus the function f(x)=|x-a| is not differentiable at x=a.