lim tan x-sinx / sin3x x--0

We have, limx→0 tan x - sin xsin3x = lim x→0 sin xcos x - sin xsin3x = lim x→0 sin x - sin x cos x cos x sin3x = limx→0 sin x 1 - cos xcos x sin3x = lim x→0 1 - cos xcos x sin2 x = lim x→0 1 - cos xcos x 1 - cos2x [as, sin2x = 1 - cos2x] = lim x→0 1 - cos xcos x 1 + cos x 1 - cos x = lim x→0 1cos x 1 + cos x =1cos (0) [1 + cos (0)] = 11 1 + 1 = 11 × 2 = 12 [as, cos 0 = 1]

lim tan x-sinx / sin³x

x--0

We have, \lim_{x \to 0} [\frac{\tan x - \sin x}{\sin^{3} x}] = \lim_{x \to 0} (\frac{\frac{\sin x}{\cos x} - \sin x}{\sin^{3} x}) = \lim_{x \to 0} (\frac{\sin x - \sin x \cos x}{\cos x \sin^{3} x}) = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{\cos x \sin^{3} x} = \lim_{x \to 0} \frac{(1 - \cos x)}{\cos x \sin^{2} x} = \lim_{x \to 0} \frac{(1 - \cos x)}{\cos x (1 - \cos^{2} x)} [a s, \sin^{2} x = 1 - \cos^{2} x] = \lim_{x \to 0} \frac{(1 - \cos x)}{\cos x (1 + \cos x) (1 - \cos x)} = \lim_{x \to 0} \frac{1}{\cos x (1 + \cos x)} = \frac{1}{\cos (0) [1 + \cos (0)]} = \frac{1}{1 (1 + 1)} = \frac{1}{1 \times 2} = \frac{1}{2} [a s, \cos 0 = 1]

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