lim sin[π(n^2+1)^(1/2)]
=lim (-1)^n*sin[π(n^2+1)^(1/2)-nπ]
=lim (-1)^n*sin[π/【(n^2+1)^(1/2)+n】]
=0。最后一个等式是无穷小量乘以有界量