求极限 lim sin pi(n^2+1)^(1/2)

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求极限 lim sin pi(n^2+1)^(1/2)

等于0

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

[(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]
...

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[(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

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