Question

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

Answer 1

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

&|2x-3|[x],\;x \geqslant1 \\\ & \sin \left( {\dfrac{{\pi x}}{2}} \right),\;x < 1\\\ \end{aligned} \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$$.