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find the value of 'a' for which the function f defined by

f(x)={a sin( pie/2) x+1 ,x < 0

{(tan x -sinx) / x³ ,x>0

is continuous at x=0

$G i v e n t h a t, f u n c t i o n i s c o n t i n u o u s a t x = 0 T h u s, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} f (x) = \lim_{x \to 0^{-}} f (x) ...... (1) N o w, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} \frac{\tan x - \sin x}{x^{3}} = \lim_{x \to 0^{+}} \frac{(\sin x / \cos x) - \sin x}{x^{3}} = \lim_{x \to 0^{+}} \frac{(\sin x) - \sin x (\cos x)}{(\cos x) x^{3}} = \lim_{x \to 0^{+}} \frac{\sin x (1 - \cos x)}{(\cos x) x^{3}} = \lim_{x \to 0^{+}} \frac{\sin x (1 - 1 + 2 \sin^{2} (x / 2))}{(\cos x) x^{3}} = \lim_{x \to 0^{+}} \frac{\sin x (2 \sin^{2} (x / 2))}{(\cos x) x^{3}} = 2 \lim_{x \to 0^{+}} (\frac{\sin x}{x}) \frac{(\sin^{2} (x / 2))}{2^{2} (\cos x) {(x / 2)}^{2}} = \frac{2}{2^{2}} \lim_{x \to 0^{+}} (\frac{\sin x}{x}) \lim_{x \to 0^{+}} {(\frac{\sin (x / 2)}{(x / 2)})}^{2} \lim_{x \to 0^{+}} (\frac{1}{\cos x}) = \frac{1}{2} \times 1 \times 1 \times 1 = \frac{1}{2} \Rightarrow \lim_{x \to 0^{+}} f (x) = \frac{1}{2} A n d, \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (a \sin \frac{π}{2} x + 1) = \lim_{x \to 0^{-}} (a \sin \frac{π}{2} x) + \lim_{x \to 0^{-}} (1) = \lim_{x \to 0^{-}} ((π / 2) a \frac{\sin \frac{π}{2}}{π / 2} x) + \lim_{x \to 0^{-}} (1) = (π / 2) a [\lim_{x \to 0^{-}} (\frac{\sin \frac{π}{2}}{π / 2} x)] + \lim_{x \to 0^{-}} (1) = (π / 2) a \times 1 + 1 = (π / 2) a + 1 \Rightarrow \lim_{x \to 0^{-}} f (x) = (π / 2) a + 1 T h u s, f r o m e q u a t i o n (1), w e h a v e, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{-}} f (x) \Rightarrow \frac{1}{2} = (π / 2) a + 1 \Rightarrow a = - \frac{1}{π} H e n c e, v a l u e o f a i s - \frac{1}{π} .$

-85

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