Question

If $f\left(x\right)=1$ for $x<0=1+\sin x$ for $0\le x<\pi /2,$ then at x=0, then show that the derivative $f^{\prime }\left(x\right)$ does not exist.

Answer 1

We have $f\left(0\right)=1+\sin 0=1.$

RHL $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{1+\sin (0+h)-1}{h}=\underset{h\to 0}{lim}\frac{\sin h}{h}=1$

and
LHL $L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{1-1}{-h}=0$

since value of RHL and LHL is not equal to the value of function at the point

Hence $f^{\prime }\left(0\right)$ does not exist.