Show that $\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0$ Show that the following limit holds
$$
\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0 
$$
for a fixed value of $k$
I really am just stuck at the first step here. Normally I would consider tackling this using L'Hopitals rule, however $\binom{n}{k}$ is not differentiable. I was considering using the binomial theorem, but that is for sums, not just the single scenario. Any help appreciated!
If you accept as known $\binom{n}{k} \leqslant \frac{n^k}{k!}$, then you obtain it by estimation.
Addition: added second part, as it can be helpful for somebody.
$$\frac{n^k}{k^k} \leqslant \binom{n}{k} \leqslant \frac{n^k}{k!}$$
Set k = n since it is the greatest it will ever be. The numerator will never surpass 1. So you simply have the ratio of 1 to infinity.