Let \(𝑓(𝑛) = 𝑛\) and \(𝑔(𝑛) = 𝑛^{(1+ sin \ 𝑛)}\) , where \(𝑛\) is a…

Key idea: sin n oscillates between -1 and 1, and for infinitely many integers n it can be made arbitrarily close to -1 or to 1.

• If sin n is very close to -1 (along some subsequence), then 1+sin n is near 0, so g(n)=n^{1+sin n}=n^{ε} with ε→0. In that case g(n)=o(n) and f(n)/g(n)→∞, so f(n) is not O(g(n)).

• If sin n is very close to 1 (along another subsequence), then 1+sin n is near 2, so g(n)≈n^2 and f(n)/g(n)≈1/n→0, so f(n) is not Ω(g(n)).

Conclusion: Because g(n) can be arbitrarily much smaller than n for some n and arbitrarily much larger for other n, neither f(n)=O(g(n)) nor f(n)=Ω(g(n)) holds. Therefore neither statement is correct.