lim(5n-根号(an^2+bn+c))=2,求实数a,b,c
人气:319 ℃ 时间:2020-05-12 14:21:18
解答
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=252=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)] ...可是C的答案是大于等于-5耶,怎么做的啊,求解释因为c在根号内 所以要保证有意义所以c>-5
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