Asymptotic order

Question

Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it. $$\frac{e^{n\log n}}n(A),n^{\sqrt n}(B),2^{n\log n}(C)$$ I want to order these functions based on increasing asymptotic order.

I rewrite A as $\frac{n^{nlog e}}{n}=\frac{n^{1.44n}}{n}=n^{1.44n-1}$

C is re-written as $n^{nlog2}=n^n$

So, finally, my order came to be A, B, C in increasing asymptotic growth.

Am I correct?

Answer 1

$$\frac{e^{n\log_2n}}n=\frac{e^{n(\ln n)/(\ln 2)}}n=\frac{n^{n/\ln 2}}n=n^{n/\ln 2-1}$$ $$2^{n\log_2 n}=n^n$$ We now compare powers: $\sqrt n<\frac n{\ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.