Question

Show that the function f(x) = $|x-3|,\,x\in R$, is continuous but not differentiable at x = 3.

Answer 1

To prove that the function f(x) = $|x-3|$ is continuous at x = 3 we will first find the left hand limit and right hand limit of the given function and prove both limits to be equal to each other. To prove that the given function is not differentiable at x = 3, we will find the left hand and right hand differentiability of the given function and observe both limits to be different from each other.

Complete step by step answer:
First, we will find both the left hand and right hand limits at x = 3 of the given function to prove that it is continuous at x = 3.
We are given that,
f(x) = $|x-3|$
let us consider the left hand side limit first,
$\begin{aligned} & \underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right) \\\ & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,|x-3| \\\ \end{aligned}$
As we approach x from left side of 3, x – 3 will be negative, hence |x – 3|, can be represented as
-(x – 3), so we get,
$\begin{aligned} & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,-(x-3) \\\ & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,-(3-3) \\\ & =0 \\\ \end{aligned}$
Hence the left hand limit at x = 3 is 0.
Now we will find the right hand limit,

& =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right) \\\ & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,|x-3| \\\ \end{aligned}$$ As we are approaching from positive side of x = 3, x – 3 will be positive hence |x – 3|, can be represented as (x – 3), so we get, $$\begin{aligned} & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\left( x-3 \right) \\\ & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\left( 3-3 \right) \\\ & =0 \\\ \end{aligned}$$ = LHL Hence both the right hand limit and right hand limit are equal function is continuous at x = 3. To prove it’s non differentiability at x = 3, We will first find it’s left hand derivative(LHD), which is, $\begin{aligned} & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 3 \right)}{x-3} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 3-h \right)-f\left( 3 \right)}{3-h-3} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3-\left( 3-h \right)-0}{3-h-3} \\\ & =-1 \\\ \end{aligned}$ Now right hand derivative(RHD) at x = 3, we get as $\begin{aligned} & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 3 \right)}{x-3} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 3+h \right)-f\left( 3 \right)}{3+h-3} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(3+h)-3-0}{3+h-3} \\\ & =1 \\\ & \ne LHD \\\ \end{aligned}$ **Hence we proved that RHD $\ne $ LHD , the given function is not differentiable at x = 3.** **Note:** You need to be cautious while finding opening the mod function while calculating limits as well as derivative because we are checking the value of function at the corner points, i.e. where it’s slope changes hence function at that point will also change i.e. mod part will change. These types of questions demand practice from us so try to solve more questions of this type.