Question

Show that the function is continuous but not differentiable at x = 1.

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}\left|2x-3\right|\left[x\right],x\ge 1\\ \sin \left(\frac{\mathrm{\pi x}}{2}\right),x<1\end{array}\right.$

Continuity at x = 1:
(LHL at x = 1) = $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(1-h)=\underset{h\to 0}{\lim }\sin \left(\frac{\mathrm{\pi }(1-h)}{2}\right)=\sin \frac{\mathrm{\pi }}{2}=1$

(RHL at x = 1) = $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(1+h)=\underset{h\to 0}{\lim }\left|2(1+h)-3\right|\left[1+h\right]=\underset{h\to 0}{\lim }\left|2(1+h)-3\right|=1$

Hence, (LHL at x = 1) = (RHL at x = 1)

Differentiability at x = 1:

$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{1-h-1} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }\left(1-h\right)}{2}}\right)-1}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{\cos {\displaystyle \frac{\mathrm{\pi }h}{2}}-1}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=-\frac{\mathrm{\pi }}{2}\underset{h\to 0}{\lim }\frac{\cos {\displaystyle \frac{\mathrm{\pi }h}{2}}-1}{{\displaystyle \frac{\mathrm{\pi }}{2}h}}=0 \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{1+h-1} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{-\left(2\left(1+h\right)-3\right)-1}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{-2h}{h}=-2 \end{gathered}$$

LHD ≠ RHD

Hence, the function is continuous but not differentiable at x = 1.