作者LuisSantos (^______^)
看板trans_math
標題Re: [微分] 93成大企管
時間Wed Oct 24 13:38:47 2007
※ 引述《bitchdog (PEaCE)》之銘言：
: Given f(x)= x^2-2x-1
: ----------- if x≠1+√2 and f(x) is continuous at x=1+√2
: x-1-√2
: Find f(1+√2)
f(x) is continuous => lim f(x) = f(1+√2)
x→(1+√2)
lim f(x)
x→(1+√2)
x^2 - 2x - 1
= lim --------------
x→(1+√2) x - 1 - √2
(x - (1+√2))(x - (1 - √2))
= lim ------------------------------
x→(1+√2) x - (1+√2)
= lim x - (1 - √2)
x→(1+√2)
= lim x - 1 + √2
x→(1+√2)
= 1 + √2 - 1 + √2 = √2 + √2 = (2)(√2)
所以 f(1 + √2) = (2)(√2)
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