Understanding an instance of upper bound, lower bound algorithmic analysis

Question

I am continuing on my task of understanding asymptotic analysis. It may be best to just have a meta post, if mods would prefer. Anyway:

I have two functions:

 f(n) = n^2
 g(n) = (log n)^80

From analysis with l'Hopitals rule:

 lim(n->∞) f(n)/g(n) = f'(n)/g'(n)

Which leaves us with us with:

 f'(n)/g'(n) = 2n/(80*(log n / √2)

Which will eventually lead us to:

 0/g''(n) = 0 

Which, as I understand it, shows that f(n) = o(g(n))

Is my understanding correct?

Answer 1

You can apply L'Hopital rule either if both of numerator and denominator converge to zero or infinity. So Your approach is true in general. But you made mistake to compute g'(n).

g'(n) = (80 * log(n)) * 1/(2 ln (n))
  => f'(n)/g'(n) = 2n / ((80 * log(n)^79) * 1/(n ln(2))) 
   = 2n^2 / 80log(n)^79 ln(2)

At the moment, the limit of f'(n)/g'(n) is ∞/∞ too. So you can apply L'Hopital rule again. But the result is same. But after 80th apply, you have this:

2^80 n^2 / 80! ln(2)^80
  =>  lim(n->∞) f(n)/g(n) = ∞ 

Thus, g(n) = o(f(n)).