Big-Omega proof using L'Hopital's Rule?

Question

Prove or disprove: $15n^2$ is in $\Omega(3 \times 2^n)$

So we'd have to prove or disprove this statement: $$ \exists c \in\mathbb{R}^+,\,\exists B\in\mathbb{N}, \forall n \in\mathbb{N}, n ≥ B \Rightarrow 15n^2 \geq c(3 \times 2^n) $$ There are other approaches to this but I'm interested in knowing how to prove it using L'Hopital's Rule and what steps to take.

Answer 1

To prove $15n^{2} \in \Omega(3 * 2^{n})$, we consider $L = \lim_{n \to \infty} \frac{15n^{2}}{3 * 2^{n}}$.

If $L = 0$, then we conclude $15n^{2} \in \omega(3 * 2^{n})$. Little-omega is a strict inequality, and so it implies Big-Omega.

Applying L'Hospital's rule, we get:

$$L = \lim_{n \to \infty} \frac{30}{3 * (ln(2))^{2} * 2^{n}} = 0$$

I applied L'Hospital's rule twice, differentiating the numerator and the denominator twice.