Prove whether $f(n)$ is $O, o, Ω, ω$ or $Θ$ of $g(n)$. $$f(n)=\log n2^n,g(n)=2^{2\log n}$$ Assume any log. given in the problem is base $2$.
Alright so here's what I've got $$f(n)=\log n2^n=\log n+\log2^n=\log n+n\log2=\log n+n$$ $$g(n)=2^{2\log n}=2^{\log n^2}=n^2$$ Now from here I can do $f(n)/g(n)$: $$\frac{\log n+n}{n^2}$$ I then applied L'Hôpital's rule, and found the derivatives. This is where it gets fuzzy for me$\ldots$ $$\frac{1/(n\ln 2)+1}{2n}$$ I do not know how to progress from here, I know I'm close, though (if I derived correctly.)
Here's a simpler way to do it: $\log n<n$ for sufficiently large $n$, so $$\frac{\log n+n}{n^2}<\frac{2n}{n^2}=\frac2n$$ and $\lim_{n\to\infty}\frac2n=0$. Thus $f(n)=o(g(n))$.