$$\lim_{n \to \infty}{n \choose k}2^{-n}$$, for a set $k \in \mathbb{N}$
I'm trying to evaluate this limit but I really don't know how to proceed when expanding the expression. I've seen people use Stirling's approximation to solve similar problems but that is not something that we have been taught so I don't think it's expected from us to solve it that way.
A fancy yet elementary way: take a look at the series
$$\sum_{n=1}^\infty \binom nk 2^{-n}$$
Let us use D'Alembert's Test to check convergence (as this is a positive series):
$$\frac{a_{n+1}}{a_n}=\frac{\binom{n+1}k 2^{-n-1}}{\binom nk2^{-n}}=\frac{2^n(n+1)!\, k!\,(n-k)!}{2^{n+1}k!\,(n+1-k)\,n! }=\frac12\cdot\frac{n+1}{n+1-k}\xrightarrow[n\to\infty]{}\frac12<1$$
We then get the above series converges and thus the series' general term sequence converges to zero.