Saturday, 29 December 2012

real analysis - Finding the limit of a sequence involving n-th roots.



We're asked to find the limit (as n goes to infinity) of the following sequence: an=3n4 6 3n2 + 92n9  4 2n3+4.



I thought that since the limit of the numerator exists, and since the limit of the denominator also exists and is non-zero, then the limit of the fraction should be 94.



But apperently, the limit of this sequence is 4 as n+.




I don't understand why my approach is incorrect, nor why the limit of the sequence is 4.


Answer



It's just 16+914+4=4 because for example 413n40=1.


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