$f(x)=\left\{\begin{array}{l}\frac{\sin x^2}{x}, & x \neq 1 \\ 0, & x=0\end{array}\right.$, then at x = 0
Answer & explanation
Correct answer: option 4
$f'(0)=\lim\limits_{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim\limits_{h \rightarrow 0} \frac{\sin h^2}{h^2}=1$
Thus f(x) is differentiable at x = 0 at hence also continuous at x = 0.