Consider the following functions  $f(n) = 3n^{(n^{1/2})}  $$g(n) = 2^{(n^{1/2}) \log n }$  // here base of $\log n$ is 2   $h(n) = n !$Which of the following is true?$ h (n)$ is $O (f (n) )  $$h (n)$ is $O (g (n)) $ $g (n)$ is not $O (f (n) )  $$ f(n)$ is $O(g (n))$I tried this way :  To check which function is large :      $3n^{(n^{1/2})}$ = taking log   $n^{1/2} \log 3n $  $g(n) = 2^{(n^{1/2}) \log n }$ by taking log :   $((n^{1/2})\log n )\log 2$ => $0.3 (n^{1/2})\log n  $$h(n) = n !$ => $n^n  $ by taking log :   $n \log n $   So now comparing all these 3 functions I got following equality :  $f(n) < g(n) < h(n)  $ So answer should be (4) but actually answer is (1).Could anyone explain me where I am wrong ? 
Relations of some exponential functions -- which is correct?
asymptotics