Prove that $(\log(n))^{100} = O(n^{0.01})$

Question

I am really having difficulty getting started with this. Please help me prove that $$(\log(n))^{100} = O(n^{0.01})$$ (Should we use induction for this?)

Answer 1

You have to prove that there is an $\bar n$ and a constant $M>0$ such that $$ \left| {\log ^{100} \left( n \right)} \right| \leqslant Mn^{0.01}, \;\forall n > \overline n $$ Since it is $$ \mathop {\lim }\limits_{n \to + \infty } \frac{{\log ^{100} \left( n \right)}} {{n^{0.01} }} = 0 $$ by definition of limit you can choose any $M>0$ and you have that there exists $\bar n$ such that for every $n>\bar n$ it is $$ \frac{{\log ^{100} \left( n \right)}} {{n^{0.01} }} $$ and you have done.