algorithm - When c > 0 Log(n) = O(n)? Not sure why it isn't O(log n) -


in homework, question asks determine asymptotic complexity of n^.99999*log(n). figured closer o( n log n) answer key suggests when c > 0, log n = o(n). i'm not quite sure why is, provide explanation?

it's true lg n = o( nk ) (in fact, o(nk); did hint that, perhaps?) any constant k, not 1. consider k=0.00001. n0.99999 lg n = o(n0.99999 n0.00001 ) = o(n). note bound not tight, since choose smaller k, it's fine n0.99999 lg n o(n0.99999 lg n), n lg n o(n lg n).


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