Let f(x) = ((π)/(6) ((π)/(2)sin x)) , for all x ∈ R and g(x) =...

$f\left(x\right) = \sin \left(\dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}sin x\right)\right)$
$\dfrac{- \pi}{2} \le \dfrac{\pi}{2}sin x \le \dfrac{\pi}{2}$
$- 1 \le \sin \left(\dfrac{\pi}{2}sin x\right) \le 1$
$\sin \left(\dfrac{- \pi}{6}\right) \lesin \left(\dfrac{\pi}{6}\left(\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right) \le sin \dfrac{\pi}{6}$
$\dfrac{- 1}{2} \le \sin \left(\dfrac{\pi}{6}\left(\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right) \le \dfrac{1}{2}$
$f\left(g\left(x\right)\right) = \sin \left(\dfrac{\pi}{6}\left(\sin \left(\dfrac{\pi}{2}\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right)\right)$
$- 1 \le \sin \left(\dfrac{\pi}{2}sin x\right) \le 1$
$\dfrac{- \pi}{2} \le \dfrac{\pi}{2}\sin \left(\dfrac{\pi}{2}sin x\right) \le \dfrac{\pi}{2}$
$- 1 \le \sin \left(\dfrac{\pi}{2}\left(\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right) \le 1$
$- \dfrac{\pi}{6} \le \dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}\left(\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right) \le \dfrac{\pi}{6}$
$\dfrac{- 1}{2} \le \sin \left(\dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}\sin \left(\dfrac{\pi}{2}sin x\right)\right)\right) \le \dfrac{1}{2}$
$\lim_{x \to 0}\dfrac{f\left(x\right)}{g\left(x\right)} = \lim_{x \to 0} \dfrac{\sin \left(\dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}\sin x\right)\right)}{\dfrac{\pi}{2}sin x} \times \dfrac{\dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}sin x\right)}{\dfrac{\pi}{6}\sin \left(\dfrac{\pi}{2}sin x\right)}$

$= \lim_{x \to 0} \dfrac{\pi}{6}\dfrac{\sin \left(\dfrac{\pi}{2}sin x\right)}{\dfrac{\pi}{2}sin x} = \dfrac{\pi}{6}$.


Range of g\left(f\left(x\right)\right)$is \left[\dfrac{\pi}{2}\sin \left(\dfrac{- 1}{2}\right), \dfrac{\pi}{2}sin \dfrac{1}{2}\right]$
$= \left[- \dfrac{\pi}{2}sin \dfrac{1}{2}, \dfrac{\pi}{2}sin \dfrac{1}{2}\right]$
Hence, $1$, does not belong to this range.