lim(5n-根号（an^2+bn+c)）=2,求实数a,b,c

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/10 12:03:42

n→ ∞
则lim [5n-√(an^2+bn+c)]/n=lim 2/n=0
则lim 5-√(an^2+bn+c)/n]=0
则√a=5,a=25
2=lim(5n-√（25n^2+bn+c)）{做分子有理化}
=lim [25n^2 -(25n^2+bn+c)]/[5n+√（25n^2+bn+c)]
=lim[-bn-c]/[5n+√（25n^2+bn+c)] {上下同时除以n}
=lim [-b-c/n]/[5+√（25+b/n+c/n^2)]]
=lim[-b]/10
所以b= -20
此时原式化为：lim [5n-√(25n^2 - 20n+c)]= 2
c为任意常数都行.
再问：可是C的答案是大于等于-5耶，怎么做的啊，求解释
再答：因为c在根号内所以要保证有意义所以c＞-5

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