哭求解！極限

lim x→0 ( cos 2x - cos 6x ) / x^2
=lim x→0 ( -2sin 2x + 6sin 6x ) / 2x
= lim x→0 ( -4cos 2x + 36cos 6x ) / 2
=(-4+36)/2
=16





2007-01-15 17:30:44 補充：
  這個符號多佐出來其實只要反覆用L'Hospital Rule 便成( cos 2x - cos 6x ) / x^2 (0/0 type)( -2sin 2x + 6sin 6x ) / 2x (0/0 type)

2007-01-15 17:36:43 補充：
if you don't know this rule, please go to herehttp://mathworld.wolfram.com/LHospitalsRule.html

2007-01-16 08:50:08 補充：
人地知道答案啦 在(23:21:09) http://hk.knowledge.yahoo.com/question/?qid=7007011504148飛天魏國大將軍張遼個答案正確無誤我唔copy個方法落去補充到﹐因為冇意思不如你轉去個邊用LHospitalsRule 做多次0甘0米double 獎賞

我都計到ans.:
=16
步驟你看前輩

除上面的方法外
可用這個
　lim x→0 ( cos 2x - cos 6x ) / x^2
= lim x→0 { -2 sin[(2x + 6x)/2] sin[(2x - 6x)/2] } / x^2
= lim x→0 [ -2 sin4x sin(-2x) ] / x^2
= lim x→0 ( 2 sin4x sin2x ) / x^2
= lim x→0 [ 2 (sin4x / x) (sin2x / x) ]
= lim x→0 [ 16 (sin4x / 4x) (sin2x / 2x) ]
= 16．1．1
= 16

公式為cosA - cosB = -2 sin [(A + B)/2] sin[(A - B)/2]

2007-01-16 17:26:34 補充：
上面的唔使咁串既首先題目冇話到表明佢已經知道這方法而我亦唔知道佢已經知道其次我見佢連這題都話哭求解所以估佢唔係讀緊預科因此認為佢未必識L'Hospital Rule硬係要人地用L'Hospital Rule都唔係好事一來人地未學到二來考試用唔著,用左都唔會有分再者此乃一網上供人問答既平台我要答人野,關你鬼事唔通答人野都要得你批准最後我答過所有學術性既問題都冇copy過人全部都係自己諗