Question

Show that the function \[f(x)=\left\{ \begin{align}
  &|2x-3|[x],\;x \geqslant1 \\
 & \sin \left( {\dfrac{{\pi x}}{2}} \right),\;x < 1\\
\end{align} \right.\]
is continuous but not differentiable at \[x = 1\].

Answer 1

Hint:In this problem we have to show that the given function is continuous but not differentiable at \[x = 1\]. We will first check for continuity at \[x = 1\] by verifying the left hand limit and the right hand limit . Then we check the differentiability at \[x = 1\] by verifying the left hand derivative and right hand derivative.

Complete step by step answer:
To check continuity at a point \[x = a\] we will verify ,
\[\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)\]
For differentiability at \[x = a\] we verify
\[\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f(x) - f(1)}}{{x - 1}} = f(0)\]
 A function \[f(x)\] is said to be continuous at a point \[x = a\] of its domain if \[\mathop {\lim }\limits_{x \to a} f(x)\] exist and equal to \[f(a)\]. Thus \[f(x)\] is continuous at a point \[x = a\] if \[\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)\].If the function is not continuous then it is said to be discontinuous at that point.
The given function is
\[f(x)=\left\{ \begin{align}
&|2x-3|[x],\;x \geqslant1 \\
& \sin \left( {\dfrac{{\pi x}}{2}} \right),\;x < 1\\
\end{align} \right.\]

First we check whether the function is continuous at \[x = 1\].
Hence we find right hand limit as
i.e. \[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 + h)\] --------(1)
since, \[f(x) = 2x - 3|[x] , x \geqslant 1\] where, \[x = (1 + h)\]
we can apply the values in equation (1), then we can get
\[\mathop {\lim }\limits_{x \to {1^ + }} f(2x - 3|[x] ) = \mathop {\lim }\limits_{h \to 0} |2(1 + h) - 3|[1 + h]\]
\[ = \mathop {\lim }\limits_{h \to 0} |2h - 1|[1 + h]\]
on solving the greatest integer function\[[1 + h] = 1\] and putting the limit into the function , we have
\[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = |2(0) - 1|(1) = 1\]
Now we find the left hand limit as follow
\[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 - h) \\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin \left( {\dfrac{{\pi (1 - h)}}{2}} \right) \]

Putting the limit into the function we get
\[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin \left( {\dfrac{{\pi (1 - 0)}}{2}} \right) = \sin \dfrac{\pi }{2} = 1 \]
Now we find the value of function at \[x = 1\].
\[f(1) = |2(1) - 3|[1] = 1\]
Hence we see that \[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1)\]
Hence, the given function is continuous at \[x = 1\].
Now we check for differentiability at \[x = 1\].
We find left hand derivative at \[x = 1\]
i.e. \[\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{(1 - h) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \dfrac{{\pi (1 - h)}}{2} - 1}}{{ - h}}\]

The above function is of zero by zero form so solving using the L hospital rule we differentiate the numerator and denominator. Thus we get
\[\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{( - 1)\cos \dfrac{{\pi (1 - h)}}{2}}}{{( - 1)}} = \mathop {\lim }\limits_{h \to 0} \cos \dfrac{{\pi (1 - h)}}{2}\]
Putting the limits we get \[\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \cos \dfrac{{\pi (1 - 0)}}{2} = 0\]
Now we find the right hand limit as follow
\[\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{{(1 + h) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{|2h - 1|[1 + h] - 1}}{h}\]
On solving by put the limits, we have
\[\therefore \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{|2h - 1|(1) - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - 2h - 1}}{h} = - 2\].

Hence we see that the right hand derivative does not equal the left hand derivative.Therefore the given function is not differentiable.

Note:Every differentiable function is continuous but not vice -versa. Greatest integer function gives the value of the integer less than equal to the given number. for example : \[[2.15] = 2\]. Modulus of function gives the positive values of the number. for example: \[| - 2| = 2\].