Evaluating $\lim_{m \to \infty} \binom{m}{k} \cdot 2^{-m}$ for $0 \leq k \leq m$. I am attempting to evaluate the limit $\lim_{m \to \infty} \binom{m}{k} \cdot 2^{-m}$ for $0 \leq k \leq m$, and I have been able to confirm via Mathematica that this limit is indeed 0. How would one actually go about proving this?

For simplicity let $m$ be even. $\binom{m}{k}$ is largest when $k=\frac12m$ [see here]. Hence $$\binom{m}{k}2^{-m}\le\frac{m!}{(m/2)!(m/2)!}2^{-m}.$$ Now using this you can find the limit to be zero.
