Let f _ 1 : ℝ → ℝ, f _ 2 (- (π)/(2), (π)/(2)) → ℝ, f _ 3 : (- 1, e...

(i) ${f}_{1}\left(x\right) = sin\sqrt{1 - {e}^{- {x}^{2}}}$. Clearly ${f}_{1} ( x )$ is continuous at $x = 0$ and $f\left(0\right) = 0$
${f}_{1}^{′}\left(0\right) = \lim_{x \to 0}\dfrac{sin\sqrt{1 - {e}^{- {x}^{2}}}}{\sqrt{1 - {e}^{- {x}^{2}}}} \cdot \sqrt{\dfrac{1 - {e}^{- {x}^{2}}}{{x}^{2}}} \cdot \dfrac{\left|x\right|}{x}$

$= 1 \cdot 1 \cdot \lim_{x \to 0}\dfrac{\left|x\right|}{x}$, which does not exist. So it is not differentiable at $x = 0$.
So, $P \to 2$
(ii) {f}_{2}\left(x\right) = \left\{\begin{cases}\dfrac{\left|\sin x\right|}{\tan^{- 1}x}, & x \ne 0 \\ 0 & x = 0\end{cases}\right)<br />\lim_{x \to {0}^{+}}\dfrac{\sin x}{x}\dfrac{x}{\tan^{- 1}x} = 1$and$\lim_{x \to {0}^{-}} - \dfrac{\sin x}{x} \times \dfrac{x}{\tan^{- 1} x} = - 1$<br />$⇒ {f}_{2}\left(x\right)$is not continuous at$x = 0$<br />So$Q \to 1$<br />(iii) In the close neighborhood of$x = 0$given function${f}_{3} ( x ) = 0$hence${f}^{′} 3 ( x ) = 0$also${f}^{′} 3 ( x )$is continuous at$x = 0$<br />So,$R \to 4$<br />(iv)${f}_{4}\left(x\right) = \left\{$\begin{cases}{x}^{2}sin \dfrac{1}{x}, & x \ne 0 \\ 0, & x = 0\end{cases}$\right)

Although ${f}_{4} ' \left(0\right) = 0$ due to sandwich theorem, but $\lim_{x \to 0}{f}_{4} ' \left(x\right)$ does not exist because $\cos \left(\dfrac{1}{x}\right)$ oscillates infinitely many times near $x = 0$.
So $S \to 3$