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lim. x->0

1- cosx.cos2x.cos3x /sin²2x

$S i n c e, \cos x \cos 2 x \cos 3 x = \frac{1}{2} (2 \cos x \cos 2 x \cos 3 x) = \frac{1}{2} [(2 \cos x \cos 2 x) \cos 3 x] = \frac{1}{2} [(\cos 3 x + \cos x) \cos 3 x] (\sin c e \cos A + \cos B = \cos (A + B) + \cos (A - B)) = \frac{1}{2} [\cos^{2} 3 x + \cos x \cos 3 x] = \frac{1}{4} [2 \cos^{2} 3 x + 2 \cos x \cos 3 x] = \frac{1}{4} [{2 \cos}^{2} 3 x + \cos 4 x + \cos 2 x] = \frac{1}{4} [1 + \cos 6 x + \cos 4 x + \cos 2 x] (\sin c e 2 \cos^{2} θ = 1 + \cos 2 θ) N o w, \lim_{x \to 0} \frac{1 - \cos x \cos 2 x \cos 3 x}{\sin^{2} 2 x} = \lim_{x \to 0} \frac{1 - \frac{1}{4} [1 + \cos 6 x + \cos 4 x + \cos 2 x]}{\sin^{2} 2 x} = \lim_{x \to 0} \frac{4 - 1 - \cos 6 x - \cos 4 x - \cos 2 x}{{4 \sin}^{2} 2 x} = \lim_{x \to 0} \frac{3 - \cos 6 x - \cos 4 x - \cos 2 x}{{4 \sin}^{2} 2 x} = \lim_{x \to 0} \frac{1 - \cos 6 x + 1 - \cos 4 x + 1 - \cos 2 x}{{4 \sin}^{2} 2 x} = \lim_{x \to 0} \frac{2 \sin^{2} 3 x + {2 \sin}^{2} 2 x + {2 \sin}^{2} x}{{4 \sin}^{2} 2 x} (\sin c e 2 \sin^{2} θ = 1 - \cos 2 θ) = \lim_{x \to 0} \frac{\frac{2 \sin^{2} 3 x}{x^{2}} + \frac{{2 \sin}^{2} 2 x}{x^{2}} + \frac{{2 \sin}^{2} x}{x^{2}}}{\frac{{4 \sin}^{2} 2 x}{x^{2}}} (D i v i d i n g n u m e r a t o r a n d d e n o \min a t o r b y x^{2}) = \lim_{x \to 0} \frac{{2 \times 9 (\frac{\sin 3 x}{3 x})}^{2} + {2 \times 4 (\frac{\sin 2 x}{2 x})}^{2} + {2 (\frac{\sin x}{x})}^{2}}{{4 \times 4 (\frac{\sin 2 x}{2 x})}^{2}} = \frac{{18 \lim_{x \to 0} (\frac{\sin 3 x}{3 x})}^{2} + {8 \lim_{x \to 0} (\frac{\sin 2 x}{2 x})}^{2} + {2 \lim_{x \to 0} (\frac{\sin x}{x})}^{2}}{{16 \lim_{x \to 0} (\frac{\sin 2 x}{2 x})}^{2}} = \frac{18 + 8 + 2}{16} = \frac{28}{16} = \frac{7}{4} H e n c e, \lim_{x \to 0} \frac{1 - \cos x \cos 2 x \cos 3 x}{\sin^{2} 2 x} = \frac{7}{4}$

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