I believe that Ln(n) is <= 2^(sqrt(2*lg(n))), so I am trying to prove it
I know that you can show Big O by showing Little O, so I am trying to take the limit as n approaches infinity of Ln(n)/2^(sqrt(2*lg(n)))
However I am finding it hard to prove this using the limit definition
I would appreciate any help to see how to solve this! Thanks
ln
on the left andlg
on the right? – Mooing Duck Mar 16 '17 at 01:422 ^ (2*lg(n)) ^ (1/2)
. Is that the same assqrt(2 ^ (2*lg(n)))
? Or2^(sqrt(2*lg(n)))
? – Mooing Duck Mar 16 '17 at 01:33lgn
? Is2lgn
the same as2*log(n)
? – Mar 16 '17 at 01:20