微積分題目

Question

我想請問lim 6x^2 (cotx) (csc2x)=?
x- 0

Answer 1

LIM (x-> 0){(6x^2) / (tanx six2x)} (this is 0/0 indeterminate form)
= LIM (x->0) { (12x) / [(sec^2 x)(sin2x) + (tanx)(cos2x)2] } by L'Hop.
= still 0/0, use L'Hopital's rule again
= 12 / (0+2+2(1+0))
= 12/4 = 3

我趕時間, 可能有誤.

2007-05-23 12:27:04 補充：
第二次用 L'Hospital's rule 後, 結果是:
Lim (x->0) { 分子為 12, 分母為
[2 secx (secx tanx) sin2x + (sec^2 x) (cos2x) 2] +
2 [ (sec^2 x) cos2x + (tanx)(-sin2x) 2]