I need to prove the following asymptotic relation for the purpose of cacluating a recurrence relation:$$n^3log^4n=O(n^4)$$I tried and failed to do it with induction, which, if possible using basic Calculus 1 level math, I would like you to help me with.I am not required to do it with induciton or anything, I just wondered if I can do it.I was able to prove it though, in a differet manner:$$f \in o(g) \Rightarrow f \in O(g)$$$$lim_{n\to \infty} \frac{n^3log^4n}{n^4}=0$$Using L'Hpital's rule 4 times, which proves that:$$n^3log^4n=o(n^4)$$By the definitino of $o$.Therefore it also follows that $$n^3log^4n=O(n^4)$$For the basis of the induction let $n=1$ and with $c=1$:$$f(1) = 1^3log^4(1) = 0 \leq 1=1^4 \cdot 1=c \cdot g(1)$$From here I think that it would be enough to show that:$$log^4(n) \leq n$$Although I am not sure how to continue from here.
Using Induction to prove that $n^3log^4n=O(n^4)$
asymptotics;recurrence relation;induction