lim(5n-根号（an^2+bn+c)）=2,求实数a,b,c

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.