# algorithm analysis: shows that \$ lg (n!) \$ is or is not \$ o ( lg (n ^ n)) \$ and \$ omega ( lg (n ^ n)) \$

My instructor assigned a problem that asks us to determine which asymptotic limits apply to a given $$f (n)$$ for a certain $$g (n)$$, in my case $$f (n) = lg (n!)$$ Y $$g (n) = lg (n ^ n)$$. For clarity, the convention that we use in our class is that $$lg = log_2$$, the "binary logarithm".

I know that by the Stirling approach, $$lg (n!)$$ grows in $$O (n lg (n))$$, and evaluating the limit $$lim_ {n to infty} frac {n lg (n)} {n lg (n)} = C$$, some constant> 0, and so $$lg (n!)$$ is in $$theta ( lg n ^ n)$$.

$$theta$$ It also means that my $$f (n)$$ It's in $$O (g (n))$$ Y $$Omega (g (n))$$, but this does not mean that my $$f (n)$$ is in $$o (g (n))$$ or $$omega (g (n))$$.

For that, I think I would have to evaluate $$lim_ {n to infty} frac { lg (n!)} { lg (n ^ n)}$$, But I'm not sure.

What strategy would you use to show that? $$f (n)$$ is in $$o (g (n))$$ or $$omega (g (n))$$? I would evaluate $$lim_ {n to infty} frac { lg (n!)} { lg (n ^ n)}$$?