Combinatorial limit

$\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k} = (1+1)^n = 2^n}$ $\large\displaystyle{\sum^{\lfloor n/3 \rfloor}_{k=0}\dbinom{n}{3k} = \dfrac{(1+1)^n+(-\omega)^n+(-\omega^2)^n}{3} = \dfrac{2^n-1-1}{3} = \dfrac{2(2^{n-1}-1)}{3}}$ $\implies 2^n \div \dfrac{2(2^{n-1}-1)}{3} = \dfrac{3 \times 2^n}{2(2^{n-1}-1)} = \dfrac{3 \cdot 2^{n-1}}{2^{n-1}-1} \\ \therefore \lim_{n \to \infty}{\dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{\lfloor n/3 \rfloor}_{k=0}\dbinom{n}{3k}} = \lim_{n \to \infty}{\dfrac{3 \times 2^{n-1}}{2^{n-1}-1}} = \boxed{3}}$

Note: Here, $\large\omega$ is a third root of unity.