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2 Answers
- Anonymous8 years agoFavorite Answer
k = (x tg 2x)/(1 - cos 2x)
lim x → 0
k = (x tg 2x)/(1 - (1 - 2 sin² x))
lim x → 0
k = (x tg 2x)/(2 sin² x)
lim x → 0
k = (x/(2 sin x)) (sin 2x/cos 2x)(1/sin x)
lim x → 0
k = (x/(2 sin x)) (2 sin x cos x/sin x)(1/cos 2x)
lim x → 0
k = (x/(2 sin x)) (2 cos x)(1/cos 2x)
lim x → 0
k = (1/2)(2)
k = 1
- capt_rafhasantiiLv 78 years ago
lim x→0 (x.tan 2x)/(1 - cos 2x)
= lim x→0 (x.tan 2x)/{1 – (1 – 2 sin² x)}
= lim x→0 (x.tan 2x)/ 2 sin² x
= lim x→0 (x.tan 2x) / (2 sin x.sin x)
= (1.2)/(2.1.1)
= 2/2
= 1
