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[求助] limit for sequence ,plz help me!!

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limit for sequence ,plz help me!!

find the limits of the sequence (assuming we already know lim 1/(n^1/2)=lim1/n=0   n-->inf):



1/((n^2-n+2)^1/3)

thank you!!
   

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lim_n-> infinity 1/((n^2-n+2))^(1/3)
=lim_n-> infinity 1/[(n-2)(n+1)]^(1/3)
=[lim_n->infinity 1/(n-2)^(1/3)][lim_n->infinity 1/(n+1)^(1/3)]
Let p=(n-2)^(1/3) and q=(n+1)^(1/3)
As n->infinity, p-> infinity and q->infinity
=(lim_p->infinity 1/p)(lim_q->infinity q)
=0

其實可以直出答案

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sorry that i dont know why  
lim_n-> infinity 1/((n^2-n+2))^(1/3)
=lim_n-> infinity 1/[(n-2)(n+1)]^(1/3)

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回覆 3# dickyso1994 的帖子

丫sor …傻左
lim_n-> infinity 1/((n^2-n+2))^(1/3)
Let (n^2-n+2)^(1/3)=p
as n->infinity, p->infinity
hence we can get the answer 0

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ohoh.... does it still work if i use sandwish rule?if so , can you explain it for me? thank you

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回覆 5# dickyso1994 的帖子

1/(n^2-n+1/4)^(1/3) > 1/(n^2-n+2)^(1/3) > 0
lim_n->infinity 1/(n^2-n+1/4)^(1/3)
=lim_n->infinity 1/(n-1/2)^(2/3)
=0
hence lim_n->infinity 1/(n^2-n+2)^(1/3) = 0

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oh...... i understand finally.... thank you!!!

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