limx->0 sin 2x(1-cos2x) /x3

limx→0sin2x1-cos2xx3Apply L'Hopital's rule, we get=limx→0-2cos2x-1cos2x+2sin22x3x2Apply L'Hopital's rule, we get=23limx→04cos2x-1sin2xx=23limx→04cos2x-1×limx→0sin2xxconsider limx→04cos2x-1=4cos(0)-1=4-1=3Consider, limx→0sin2xxApply L'Hopital's rule, we getlimx→02cos2x1=2cos(0)=2So, 23limx→04cos2x-1×limx→0sin2xx=23×3×2=4

lim_x->0 sin 2x(1-cos2x) /x³

\lim_{x \to 0} \frac{\sin 2 x (1 - \cos 2 x)}{x^{3}} Apply L' Hopital' s rule, we get = \lim_{x \to 0} \frac{[- 2 (\cos 2 x - 1) \cos 2 x + 2 \sin^{2} 2 x]}{3 x^{2}} Apply L' Hopital' s rule, we get = \frac{2}{3} \lim_{x \to 0} \frac{(4 \cos 2 x - 1) \sin 2 x}{x} = \frac{2}{3} [\lim_{x \to 0} 4 \cos 2 x - 1 \times \lim_{x \to 0} \frac{\sin 2 x}{x}] consider . \lim_{x \to 0} 4 \cos 2 x - 1 = 4 \cos (0) - 1 = 4 - 1 = 3 Consider, \lim_{x \to 0} \frac{\sin 2 x}{x} Apply L' Hopital' s rule, we get \lim_{x \to 0} \frac{2 \cos 2 x}{1} = 2 \cos (0) = 2 So, \frac{2}{3} [\lim_{x \to 0} 4 \cos 2 x - 1 \times \lim_{x \to 0} \frac{\sin 2 x}{x}] = \frac{2}{3} \times (3 \times 2) = 4

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