※ 引述《wargod11 (哈哈哈~~)》之銘言:
: lim {√(x^2 - x + 1) - ax - b} = 0 則a,b=? (a,b是實數)
: x→∞
如果 a < = 0 , 則 √(x^2 - x + 1) - ax - b →∞
因此 a > 0 .
對足夠大的 x ,
sqrt (x^2-x+1) - (ax+b) = sqrt(x^2 - x + 1) - sqrt(a^2x^2 + 2abx + b^2)
(1-a^2)x^2 -(1+2ab)x + (1-b^2)
= --------------------------------------------------
sqrt(x^2 - x + 1) + sqrt(a^2x^2 + 2abx + b^2)
=> a^2 = 1 => a = 1
a = 1 => 1+2b = 0 => b = -1/2
有錯請指教
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◆ From: 122.127.114.241
※ 編輯: Eliphalet 來自: 122.127.114.241 (03/30 07:07)