Re: 極限問題
※ 引述《klsh520 (快大二還沒車...)》之銘言：
: -Cos2x+aSinx+b
: l i m ----------------- 之極限存在，求a和b？
: x→π/2 (x-π/2)^4
因為極限存在
π π
所以 -cos((2)(---)) + a(sin(---)) + b = 0
2 2
-cos(π) + (a)(1) + b = 0
-(-1) + a + b = 0 => a + b + 1 = 0 => a + b = -1 ------(1)
-cos2x + (a)(sinx) + b
lim ------------------------
x→π/2 (x - π/2)^4
(2)(sin2x) + (a)(cosx)
= lim -------------------------
x→π/2 (4)((x - π/2)^3)
(4)(cos2x) - (a)(sinx)
= lim ------------------------
x→π/2 (12)((x - π/2)^2)
因為極限存在
π π
所以 (4)(cos((2)(---))) - (a)(sin(---)) = 0
2 2
(4)(cos(π)) - a = 0
a = (4)(cos(π)) = (4)(-1) = -4 代入(1)得
-4 + b = -1 => b = -1 + 4 = 3
--
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