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Question-11416




Question Number 11416 by anisa last updated on 25/Mar/17
Answered by Joel576 last updated on 25/Mar/17
= lim_(x→0)  ((2 sin^2  x)/(2x sin 2x))  = lim_(x→0)  (2/2) . ((sin x)/x) . ((sin x)/(sin 2x))  = 1 . 1 . (1/2)  = (1/2)
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}}{\mathrm{2}}\:.\:\frac{\mathrm{sin}\:{x}}{{x}}\:.\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$=\:\mathrm{1}\:.\:\mathrm{1}\:.\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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