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
[(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
所以 lim
lim
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