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lim(n->无穷)[(3n^2+cn+1)/(an^2+bn)-4n]=5
求常数a、b、c
人气:113 ℃ 时间:2020-03-28 23:45:12
解答
lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5lim {[(3n^2+cn+1)-4n(an^2+bn)]/(an^2+bn)}=5lim {[-4an^3+(3-4b)n^2+cn+1]/(an^2+bn)}=5所以-4a=03-4b=0c/b=5解得:a=0 b=3/4 c=15/4
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