数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 23:31:22

数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
数列极限
lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?

数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5
lim {[(3n^2+cn+1)-4n(an^2+bn)]/(an^2+bn)}=5
lim {[-4an^3+(3-4b)n^2+cn+1]/(an^2+bn)}=5
所以
-4a=0
3-4b=0
c/b=5
解得:
a=0
b=3/4
c=15/4

a=0
b=3/4
c=15/4

此题有问题
因为lim [(3n^2+cn+1)/(an^2+bn)]=3/a
而 lim (-4n)不存在