※ 引述《LKK (衝吧!!!)》之銘言：
: sin x - tan^{-1} x
: lim ---------------------- =?
: x->0 x ln(1+x)
: 似乎不能用羅必達法則, 請問有什麼比較好方法, 謝謝
: 答: 0
泰勒展開:
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 - ...
取極限x→0，可得極限值為0
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