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

Hint: 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,
\[\Rightarrow f\left( x \right)=\left\{ \begin{align}
  & \begin{matrix}
   -x, & x \\
\end{matrix}\le 0 \\
 & \begin{matrix}
   x, & x>0 \\
\end{matrix} \\
\end{align} \right.\]
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) \[L.H.L=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)\]
\[\begin{align}
  & \Rightarrow L.H.L=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x \right) \\
 & \Rightarrow L.H.L=-\left( 0 \right) \\
 & \Rightarrow L.H.L=0 \\
\end{align}\]
(ii) \[R.H.L=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)\]
\[\begin{align}
  & \Rightarrow R.H.L=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( x \right) \\
 & \Rightarrow R.H.L=0 \\
\end{align}\]
(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,
\[\Rightarrow f\left( x \right)=\left\{ \begin{align}
  & \begin{matrix}
   -x, & x \\
\end{matrix}\le 0 \\
 & \begin{matrix}
   x, & x>0 \\
\end{matrix} \\
\end{align} \right.\]
Differentiating both the sides with respect to x, we get,
\[\Rightarrow f'\left( x \right)=\left\{ \begin{align}
  & \begin{matrix}
   -1, & x \\
\end{matrix}<0 \\
 & \begin{matrix}
   1, & x>0 \\
\end{matrix} \\
\end{align} \right.\]
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, \[f'\left( x \right)=\left\{ \begin{align}
  & \begin{matrix}
   -1, & x \\
\end{matrix}<0 \\
 & \begin{matrix}
   1, & x>0 \\
\end{matrix} \\
\end{align} \right.\], we have,
(i) \[L.H.D.=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f'\left( x \right)\]
\[\begin{align}
  & \Rightarrow L.H.D.=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -1 \right) \\
 & \Rightarrow L.H.D.=-1 \\
\end{align}\]
(ii) \[R.H.D.=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f'\left( x \right)\]
\[\begin{align}
  & \Rightarrow R.H.D.=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( 1 \right) \\
 & \Rightarrow R.H.D.=1 \\
\end{align}\]
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.