solve for a,b if
lim[x->0]
[x(1+acosx)-sinx]/(x^3)=1
limit3
2009-03-23 7:50 am
回答 (2)
2009-03-23 7:54 am
2009-03-23 8:33 am
原式= lim(x->0) { x[1+a(1 - x^2/2+...)] - b(x-x^3/6+...) }/x^3
= lim(x->0) [(1+a-b)x + (b/6 - a/2)x^3 + ... ]/x^3 = 1
=> a-b= -1, b/6- a/2 = 1
=> a= -5/2 , b= -3/2
= lim(x->0) [(1+a-b)x + (b/6 - a/2)x^3 + ... ]/x^3 = 1
=> a-b= -1, b/6- a/2 = 1
=> a= -5/2 , b= -3/2
收錄日期: 2021-04-22 00:42:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090322000051KK02229