If f x = log e 1 + x 2 ( tan x ) sin x 3 , x ≠ 0 is to be continuous at x = 0 , then f ( 0 ) must be equal to

Step-by-step solution

f x = log e 1 + x 2 ( tan x ) sin x 3 , x ≠ 0 f 0 = log e 1 + 0 tan 0 sin 0 = 0 0 Apply L'Hospital rule d uv d x = uv ' + vu ' ⇒ f 0 = lim x → 0 1 1 + x 2 tan x x 2 sec 2 x + 2 x tan x cos x 3 × 3 x 2 ⇒ f 0 = lim x → 0 ( x 1 + x 2 tan x ) ( x sec 2 x + 2 tan x ) cos x 3 × 3 x 2 = lim x → 0 x s e c 2 x + 2 tan x 3 x cos x 3 1 + x 2 tan x = lim x → 0 2 tan x x + s e c 2 x 3 cos x 3 1 + x 2 tan x ∵ lim x → 0 tan x x = 1 = lim x → 0 2 × 1 +