Show that $\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0$

Question

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!

Answer 1

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. $$\frac{n^k}{k^k} \leqslant \binom{n}{k} \leqslant \frac{n^k}{k!}$$

Answer 2

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.