Can't find lower bound for function.

Question

I am looking to find the lower bound for a function $f(n) = \frac{n^5}{n^3 + 3n}$ where $g(n) = n^2$. I am looking to prove that $f(n)$ is $\theta(n^2)$ and have already proved that $f(n) = O(n^2)$ but cannot find the lower bound for this function and prove $f(n)=\Omega(n^2)$.

Answer 1

For $n\ge 1$ we have

$$\frac{f(n)}{n^2}=\frac{n^3}{n^3+3n}\ge\frac{n^3}{(n+1)^3}=\left(\frac{n}{n+1}\right)^3\ge\frac18\,.$$

Answer 2

$f(n) = \underbrace{n^2}_{g(n)} \frac{n^3}{n^3} \frac{1}{1+3n^{-2}},$ so it is enough to bound $\frac{1}{1+3n^{-2}}$. \begin{align*} 1 &\leq n < \infty \\ 1 = 1^2 &\leq n^2 < \infty \\ 1 &\geq n^{-2} > 0 \\ 3 &\geq 3n^{-2} > 0 \\ 4 &\geq 1+3n^{-2} > 1 \\ \frac{1}{4} &\leq \frac{1}{1+3n^{-2}} < \frac{1}{1} = 1 \text{.} \end{align*} Therefore, $$ \frac{1}{4} g(n) \leq f(n) < g(n) \text{.} $$ (There's nothing special about "$1 \leq {}$" in the first line. Pick another starting point or decide to have a strict inequality. If you choose a larger lower bound or a strict inequality, you just raise the lower bound you end up with.)