f(n)=4n 2 +5n 2 ∗log(n) , g(n) = n^3g(n)=n 3 Now to show f(n)=O(g(n))f(n)=O(g(n)) when number of operation increases i.e., n\rightarrow \inftyn→∞ then \frac{f(n)}{g(n)}=\frac{4n^3+5n^2.log\ n}{n^3}=4+\frac{5.log\ n}{n} g(n) f(n) ​ = n 3 4n 3 +5n 2 .log n ​ =4+ n 5.log n ​ Now, \lim_{n\to\infty}lim n→∞ ​ \frac{f(n)}{g(n)}=4+\lim_{n\to\infty}\frac{5.log\ n}{n} g(n) f(n) ​ =4+lim n→∞ ​ n 5.log n ​ =4+5\lim_{n\to\infty} \frac{1/n}{1}=4+5lim n→∞ ​ 1 1/n ​ [Using L'Hospital Rule] =4+5\times0=4=4+5×0=4 So, |\frac{f(n)}{g(n)}|\rightarrow 4∣ g(n) f(n) ​ ∣→4 as n\rightarrow\inftyn→∞ So, f(n)=O(g(n))f(n)=O(g(n))