※ 引述《LuisSantos (^______^)》之銘言：
: ※ 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
回應 klsh520:為什麼極限存在就是=0？分母部分不用管他
f(x)
若 lim ------ = L(存在) 且 lim g(x) = 0 => lim f(x) = 0
x->a g(x) x->a x->a
f(x)
proof: ∵ lim f(x) = lim [------ * g(x)]
x->a x->a g(x)
f(x)
= lim ------ * lim g(x)
x->a g(x) x->a
= L * 0 = 0
∴ lim f(x) = 0
x->a
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