VITEEE 2018 Paper

For function to be continuous : f(0 + h) = f(0 - h) = f(0) f(0 + h) = $\lim\limits_{h\to 0}$ h sin 1/h = 0 × (a finite quantity) = 0 f(0 - h) = $\lim\limits_{h\to 0}$ - h sin (1/-h) = 0 × (a finite quantity) = 0 Also , $\lim\limits_{x\to 0}$ x sin 1/x = 0 × (a finite quantity) = 0 ⇒ function is continuous at x = 0 For function to be differentiable : f (0 + h)= f'(0-h) f ' (0 + h) = $\frac{f'(0+h)-f(0)}{h}$ = $\lim\limits_{h\to 0}$ $\frac{h \sin\frac{1}{h} - 0}{h}$ = $\lim\limits_{h\to 0}$ sin $\left(\frac{1}{h}\right)$ which does not exist