推 tn383706:謝謝你 114.38.7.29 01/29 18:20
※ 引述《tn383706 (扇子大少爺)》之銘言：
: http://www.lib.ntu.edu.tw/exam/undergra/83/83019.htm
: 2.
: lim [(√2x^2+3x+4)-ax-q] = 0 (a,q是常數)
: x→∞
: 問q=?
: 謝謝各位
sqrt( 2x^2 + 3x + 4 ) - ( ax + q )
= x ( sqrt( 2 + 3/x + 4/x^2) - a + q/x ) → 0 as x→∞
=> a = √2 .
sqrt( 2x^2 + 3x + 4 ) - ( √2 x + q )
3x + 4 3√2
= ------------------------------------------- - q → ------ - q
sqrt( 2x^2 + 3x + 4 ) + sqrt( 2x^2 ) 4
as x→∞
3√2
=> q = ------
4
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