求极限 lim sin pi(n^2+1)^(1/2)

网友回答

等于0======以下答案可供参考======
供参考答案1:
这个极限应该发散吧，n趋于无穷时(n^2+1)^(1/2)也趋向于无穷，但lim sin pi x 的极限不存在
供参考答案2:
[(n^2+1)^(1/2) - n] = [(n^2+1)^(1/2) - n] [(n^2+1)^(1/2) +n] / [(n^2+1)^(1/2) + n]
= [n^2+1-n^2 ]/ [(n^2+1)^(1/2) + n]
= 1/[(n^2+1)^(1/2) + n]
lim [(n^2+1)^(1/2) - n] = lim 1/[(n^2+1)^(1/2) + n]=0
所以 lim (n^2+1)^(1/2) = n (整数)
lim sin pi(n^2+1)^(1/2)= lim sin pi(整数) = 0