Question

lim(根号(n^2-1)-an)=0 求a的值

Answer 1

∵若a≤0,lim(n->∞)[√(n^2-1)-an]=∞

这与lim(n->∞)[√(n^2-1)-an]=0矛盾

∴a>0

∵lim(n->∞)[√(n^2-1)-an]=lim(n->∞)[((1-a^2)n^2-1)/(√(n^2-1)+an)] (有理化分子,并化简)

又lim(n->∞)[√(n^2-1)-an]=0

∴必须1-a^2=0 (∵若1-a^2≠0,lim(n->∞)[((1-a^2)n^2-1)/(√(n^2-1)+an)]=∞)

==>a^2=1

故a=1.