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I found an answer but I'm not sure if it is correct:

Let $f(n)=(2+\sin n)\cdot n$ and $g(n)=\ln n$. $0\leq g(n)\leq f(n)$ $\forall n\gt0$ so $f(n)=\Omega(g(n))$. But $f(n)\neq\Theta(g(n))$ because $f(n)$ will always grow faster than $g(n)$ regardless of a constant multiple. Additionally $f(n)\neq\omega(g(n))$ because $\lim_{n\to\infty}\frac{(2+\sin n)\cdot n}{\ln n}$ diverges and $7\ln4\gt(2+\sin4)\cdot4$.

Would these functions meet the aforementioned requirements?

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    $\begingroup$ Your limit here is infinity, is that really what you meant? $\endgroup$
    – Aphelli
    Oct 7, 2019 at 19:05
  • $\begingroup$ No, I was looking for a limit that does not exist (even in the sense of being infinite) such as $f(n)=(1+\sin n)\cdot n$ and $g(n)=n$, $\lim_{n\to\infty}\frac{f(n)}{g(n)}=\lim_{n\to\infty}(1+\sin n)$ whose limit does not exist. $\endgroup$
    – Revoltechs
    Oct 7, 2019 at 19:09
  • $\begingroup$ Then your example does not work, since $g=o(f)$. A suggestion: you could make $g(n)=\sqrt{n} + (1-(-1)^n)n$. $\endgroup$
    – Aphelli
    Oct 7, 2019 at 21:41

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