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- LopezLv 77 年前最佳解答
lim t→0 (tan2t/sin2t -1)
= lim t→0 ﹝(sin2t/cos2t)/sin2t -1﹞
= = lim t→0 (1/cos2t) -1)
= 1/cos0 -1
= 1/1 -1
= 0
Ans: 0
2013-10-20 15:17:11 補充：
tan2t/sin2t-1,考慮運算規則,先乘除後加減,分母是sin2t
但是如果分母如意見欄所述,是sin2t-1,用打字的就要列成:
lim t→0 ﹝tan2t/(sin2t-1)﹞
(原來的題目中,除號是長長的橫躺著,所以沒問題)
2013-10-20 15:35:27 補充：
lim t→0+ ﹝tan2t/(sin2t-1)﹞
= lim t→0+ (sin2t/cos2t) /﹝sin2t (1-1/sin2t)﹞
= lim t→0+ (1/cos2t) / (1-1/sin2t)
= (1/1) / (1-∞)
= 1/(-∞)
= 0
lim t→0- ﹝tan2t/(sin2t-1)﹞
= lim t→0- (1/cos2t) / (1-1/sin2t)
= (1/1) / (1+∞)
= 1/∞
= 0
故lim t→0 ﹝tan2t/(sin2t-1)﹞ = 0
Ans: 0
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