Proof/derivation of $\lim\limits_{n\to\infty}{\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\frac{an+bk}{cn+dk}}\stackrel?=\frac{2a+b}{2c+d}$?
The reason this is true is that the binomial coefficients are strongly concentrated around the mean, when $k \sim \frac{n}{2}$. Using some standard concentration inequalities (Chernoff is strong, but Chebyshev's inequality sufficies too), you can show that for any constant $c > 0$, $2^{-n} \sum_{ k \in [0, (1/2 - c ) n ] \cup [(1/2 + c)n, n]} \binom{n}{k} \rightarrow 0$ as $n \rightarrow \infty$.
Hence in your limit, the only terms that survive are when $k \sim \frac{n}{2}$, in which case you can cancel the $n$ throughout and get the right-hand side.
For the limit to hold, you clearly need $2c + d \neq 0$. You would also need $c \neq 0$, as otherwise the $k = 0$ term of the sum will present difficulties.
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} = \lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}\pars{an + bk}\int_{0}^{1}t^{cn + dk - 1}\,\,\,\dd t \\[3mm] = &\ \lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{% an\sum_{k = 0}^{n}{n \choose k}\pars{t^{d}}^{k} + b\sum_{k = 0}^{n}{n \choose k}k\pars{t^{d}}^{k}}\dd t \end{align}
However, $\ds{\sum_{k = 0}^{n}{n \choose k}\xi^{k} = \pars{1 + \xi}^{n} \quad\imp\quad \sum_{k = 0}^{n}{n \choose k}k\,\xi^{k} = n\xi\pars{1 + \xi}^{n - 1}}$.
Then, \begin{align} &\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} = \\[3mm] = &\ \lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{% an\pars{1 + t^{d}}^{n} + bn\,t^{d}\pars{1 + t^{d}}^{n - 1}}\dd t \\[3mm] = &\ \color{#f00}{\lim_{n \to \infty}\braces{\vphantom{\LARGE A}% {n \over 2^{n}}\bracks{\vphantom{\Large A}a\,\mathrm{f}_{n}\pars{cn,d} + b\,\mathrm{f}_{n - 1}\pars{cn + d,d}}}} \end{align}
$$ \mbox{where}\quad \begin{array}{|c|}\hline\\ \ds{\quad\mathrm{f}_{n}\pars{\mu,\nu} \equiv \int_{0}^{1}t^{\mu - 1}\pars{1 + t^{\nu}}^{n}\,\dd t\quad} \\ \\ \hline \end{array} $$
Can you take it from here ?. Maybe, an asymptotic study of the integral could be somehow reasonable. Otherwise, the integral is related to hypergeometric functions.
I believe a probabilistic argument can be given here.
Let $X_1, X_2, \ldots $ be iid Bernoulli random variables with success probability $p$. Then by Strong Law of Large Numbers $$\bar{X}_n:=\frac{1}n\sum_{i=1}^n X_i \stackrel{a.s.}{\to} p$$ Also for any continuous function $g$ we have $g(\bar{X}_n) \stackrel{a.s.}{\to} g(p)$. Now if $g$ is such that there exist a random variable $Y$ with finite expectation such that $g(\bar{X}_n) \le Y$ almost surely. Then by DCT we have $E(g(\bar{X}_n)) \to g(p)$.
Now this has lot of applications.
Example 1: Take $p=\frac12$. $g(x)=\frac{a+bx}{c+dx}$. Make sure $g$ is continuous $[0,1]$. Then clearly $g$ is bounded. Hence by DCT $$E(g(\bar{X}_n))=\sum_{i=1}^n \frac{\binom{n}{k}}{2^n}g\left(\frac{k}{n}\right) \to g(\frac12)=\frac{2a+b}{2c+d}$$
Example 2: Take $p=\frac12$,$g(x)=e^{-x^2}$. Clearly $g$ is continuous and bounded. Hence $$\frac1{2^n}\sum_{i=1}^n \binom{n}{k}e^{-k^2/n^2} \to e^{-1/4}$$
Similarly we can obtain plethora of such results.