Question

How do you prove that the function $f\left( x \right)=\left| x \right|$ is continuous at x = 0, but not differentiable at x = 0?

Answer 1

Remove the modulus sign by considering the situations $x\le 0$ and x > 0. Now, find the L.H.L (Left hand limit), R.H.L. (Right hand limit) and f (0). If these three values are equal then we can say that the function $f\left( x \right)$ is continuous at x = 0. Now, find $f'\left( x \right)$ by differentiating the function $f\left( x \right)$ and check the L.H.D. (Left hand derivative) and R.H.D. (Right hand derivative). If L.H.D. = R.H.D. then the function $f\left( x \right)$ is differentiable at x = 0 otherwise not.

Complete step by step answer:
Here, we have been provided with the function $f\left( x \right)=\left| x \right|$ and we have been asked to prove that it is continuous at x = 0, but not differentiable at x = 0.
We have the function $f\left( x \right)=\left| x \right|$, so removing the modulus sign by considering the cases $x\le 0$ and x > 0, we have,

& \begin{matrix} -x, & x \\\ \end{matrix}\le 0 \\\ & \begin{matrix} x, & x>0 \\\ \end{matrix} \\\ \end{aligned} \right.$$ Now, we know that for a function to be continuous we must have the value of L.H.L. (Left hand limit), R.H.L. (Right hand limit) and $$f\left( 0 \right)$$ equal. So, we must have, $$\Rightarrow L.H.L.=R.H.L.=f\left( 0 \right)$$ Let us find the three values one – by – one. So, we have, (i) $$L.H.L=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$$ $$\begin{aligned} & \Rightarrow L.H.L=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x \right) \\\ & \Rightarrow L.H.L=-\left( 0 \right) \\\ & \Rightarrow L.H.L=0 \\\ \end{aligned}$$ (ii) $$R.H.L=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$$ $$\begin{aligned} & \Rightarrow R.H.L=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( x \right) \\\ & \Rightarrow R.H.L=0 \\\ \end{aligned}$$ (iii) $$f\left( 0 \right)=-\left( 0 \right)$$ $$\Rightarrow f\left( 0 \right)=0$$ Clearly, we can see that we have L.H.L. = R.H.L. = f (0), so we can say that the function $$f\left( x \right)=\left| x \right|$$ is continuous at x = 0. Now, let us check the differentiability of the function $$f\left( x \right)$$. We have the function, $$\Rightarrow f\left( x \right)=\left\\{ \begin{aligned} & \begin{matrix} -x, & x \\\ \end{matrix}\le 0 \\\ & \begin{matrix} x, & x>0 \\\ \end{matrix} \\\ \end{aligned} \right.$$ Differentiating both the sides with respect to x, we get, $$\Rightarrow f'\left( x \right)=\left\\{ \begin{aligned} & \begin{matrix} -1, & x \\\ \end{matrix}<0 \\\ & \begin{matrix} 1, & x>0 \\\ \end{matrix} \\\ \end{aligned} \right.$$ Now, we know that a function is differentiable at a given point if and only if it satisfies the condition L.H.D. (Left hand derivative) = R.H.D. (Right hand derivative). Mathematically, at point x = 0, we must have, $$\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f'\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f'\left( x \right)$$ For the function, $$f'\left( x \right)=\left\\{ \begin{aligned} & \begin{matrix} -1, & x \\\ \end{matrix}<0 \\\ & \begin{matrix} 1, & x>0 \\\ \end{matrix} \\\ \end{aligned} \right.$$, we have, (i) $$L.H.D.=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f'\left( x \right)$$ $$\begin{aligned} & \Rightarrow L.H.D.=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -1 \right) \\\ & \Rightarrow L.H.D.=-1 \\\ \end{aligned}$$ (ii) $$R.H.D.=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f'\left( x \right)$$ $$\begin{aligned} & \Rightarrow R.H.D.=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( 1 \right) \\\ & \Rightarrow R.H.D.=1 \\\ \end{aligned}$$ Clearly, we can see that L.H.D. $$\ne $$ R.H.D. So, we can conclude that the given function $$f\left( x \right)=\left| x \right|$$ is not differentiable at x = 0. Hence, from the above results obtained it is clear that $$f\left( x \right)$$ is continuous at x = 0, but not differentiable at x = 0. **Note:** One must remember the condition for a function to be continuous and differentiable. Always remember that if a function is differentiable then it will always be continuous but the converse is not true. If a function is continuous then it may be differentiable or not. We have to check it using the rules and formula just like we did in the above solution. If a function is not continuous at some point x = a then it will never be differentiable at x = a. In the above question the point of suspicion is only x = 0 and that is why we have checked the result for only one point. If there were more than one point then we would have checked for all the points one – by – one.