f (x) = The given function f is defined for all x ∊ R. It is known that a function f is continuous at x = 0, if $\lim\limits_{x\to 0^{-}}$ f (x) = $\lim\limits_{x\to 0^{+}}$ f (x) = f (0) $\lim\limits_{x\to 0^{-}}$ f (x) = $\lim\limits_{x\to 0} \left[ \sin \frac{\pi}{2} (x+1) \right]$ = a sin $\frac{\pi}{2}$ = a (1) = a $\lim\limits_{x\to 0^{+}}$ f (x) = $\lim\limits_{x\to 0}$ $\frac{\tan x - \sin x}{x^3}$ = $\lim\limits_{x\to 0}$ $\frac{ \frac{\sin x}{\cos x} - \sin x }{x^3}$ = $\lim\limits_{x\to 0}$ $\frac{ \sin x (1 - \cos x) }{ x^3 \cos x }$ = $\lim\limits_{x\to 0}$ $\frac{ \sin x \cdot 2 \sin^2 \frac{x}{2} }{ x^3 \cos x }$ = 2 $\lim\limits_{x\to 0}$ $\frac{1}{\cos x}$ × $\lim\limits_{x\to 0}$ $\frac{\sin x}{x}$ × $\lim\limits_{x\to 0}$ $\left[ \frac{ \sin \frac{x}{2} }{ x } \right]^2_0$ = 2 × 1 × 1 × $\frac{1}{4}$ × $\lim\limits_{x/2 \to 0}$ $\left[ \frac{ \sin \frac{x}{2} }{ \frac{x}{2} } \right]^2$ = 2 × 1 × 1 × $\frac{1}{4}$ × 1 = $\frac{1}{2}$ Now, f(0) = a sin $\frac{\pi}{2}$ (0 + 1) = a sin $\frac{\pi}{2}$ = a × 1 = a Since f is continuous at x = 0, a = $\frac{1}{2}$