作者iamhido (.....................)
看板trans_math
标题Re: [微分] 请教一题极限
时间Fri Oct 29 00:08:16 2004
※ 引述《Acrylates (金色狂风(别倒着念))》之铭言：
: lim {√(x^2 - x + 1) - ax - b} = 0 则a,b=? (a,b是实数)
: x→∞
√(x^2 - x + 1) + (ax + b)
原式乘上 ---------------------------
√(x^2 - x + 1) + (ax + b)
(x^2 - x + 1) - (ax + b)^2
则原式= lim --------------------------
x→∞ √(x^2 - x + 1) + (ax + b)
(1 - a^2)x^2 - (1 + 2ab)x + (1-b^2)
= lim -----------------------------------
x→∞ √(x^2 - x + 1) + (ax + b)
因为原式极限值为0
故 (1 - a^2)x^2 = 0 => a= 1,-1
(1 + 2ab)x = 0 => b= -1/2 when a=1;b= 1/2 when a= -1
Ans: if a= 1 => b= -1/2
if a= -1 => b= 1/2
参考看看罗!!有错请指教...
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