lim_(x→0) ((√(2−2cos 2x))/(3∣sin 4x∣)) =?


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Question Number 121822 by liberty last updated on 12/Nov/20

$\lim_{x \to 0} \frac{\sqrt{2 - 2 \cos 2 x}}{3 ∣ \sin 4 x ∣} = ?$
	Answered by bemath last updated on 12/Nov/20
	${ ((lim_(x→0^+ ) ((√(2−2(1−2sin^2 x)))/(3∣sin 4x∣)) = lim_(x→0^+ ) ((2∣sin x∣)/(3∣sin 4x∣))= (1/6))),((lim_(x→0^− ) ((−2sin x)/(−3sin 4x)) = (1/6))) :} so the value of lim_(x→0) ((√(2−2cos 2x))/(3∣sin 4x∣)) = (1/6). ⧫$
	${\begin{cases} \lim_{x \to 0^{+}} \frac{\sqrt{2 - 2 (1 - 2 \sin^{2} x)}}{3 ∣ \sin 4 x ∣} = \lim_{x \to 0^{+}} \frac{2 ∣ \sin x ∣}{3 ∣ \sin 4 x ∣} = \frac{1}{6} \\ \lim_{x \to 0^{-}} \frac{- 2 \sin x}{- 3 \sin 4 x} = \frac{1}{6} \end{cases}$ $s o t h e v a l u e o f \lim_{x \to 0} \frac{\sqrt{2 - 2 \cos 2 x}}{3 ∣ \sin 4 x ∣} = \frac{1}{6} . ⧫$
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