lim
x -> 0
(x - sin[x]) * (cos[x] - 1 + x^2 / 2)
-------------------------------------------
(sec[x] - 1 - x^2 / 2) * (x - tan[x])
without using L'Hopital's rule.
thanks!
一陣風 寫到:分兩個情形來討論:x > 0 且 x -> 0,x < 0 且 x -> 0.
1. x > 0 且 x -> 0:
此時[sinx] = [cosx] = [tanx] = 0,[secx] = 1,代入原式可得極限.
2. x < 0 且 x -> 0:
此時 [sinx] = [tanx] = -1,[cosx] = 0,[secx] = 1,代入原式可得極限.