※ 引述《LKK (冲吧!!!)》之铭言： : sin x - tan^{-1} x : lim ---------------------- =? : x->0 x ln(1+x) : 似乎不能用罗必达法则, 请问有什麽比较好方法, 谢谢 : 答: 0 泰勒展开: sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - (1/7!)x^7 + ... arctan(x) = x - (1/3)x^3 + (1/5)x^5 - (1/7)x^7 + ... ln(1+x) = x - (1/2)x^2 + (1/3)x^3 - (1/4)x^4 + (1/5)x^5 - ... [(1/3)-(1/3!)]x^3 - [(1/5)-(1/5!)]x^5 + [(1/7)-(1/7!)]x^7 - ... 原式 = ───────────────────────────────── x^2 - (1/2)x^3 + (1/3)x^4 - (1/4)x^5 + (1/5)x^6 - ... 同除 [(1/3)-(1/3!)]x - [(1/5)-(1/5!)]x^3 + [(1/7)-(1/7!)]x^5 - ... = ──────────────────────────────── x^2 1 - (1/2)x + (1/3)x^2 - (1/4)x^3 + (1/5)x^4 - ... 取极限x→0，可得极限值为0 # --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 140.112.211.87