Determine $\lim\limits_{n \to \infty}{{n} \choose {\frac{n}{2}}}\frac{1}{2^n}$, where each $n$ is even

Here's another approach! Write $n=2m:$

$$\binom{2m}{m}\frac{1}{4^m}=\frac{1}{\pi}\int_0^1 t^{-\frac{1}{2}}(1-t)^{m-\frac{1}{2}}dt$$

The integrand tends to $0$ as $m\to \infty$

$$\left[ {n\choose n/2}{1\over 2^n}\right]^2={1\over n+1}\prod_{j=1}^{n/2}\left(1-{1\over 4j^2}\right) \leq{1\over n+1}\to 0.$$

This number is $0$ as you have written it.

There is the deMoivre-Laplace limit theorem showing that biased coin tosses approach a bell curve in distribution.

\[ \mathbb{P}[k \text{ heads}]= \binom{n}{k} p^k q^{n-k} \to \frac{1}{\sqrt{2\pi n pq} } e^{-(k - np)^2/2npq} \]

In our case, $p = 1/2, k = n/2$

\[ \mathbb{P}[n/2 \text{ heads}]= \binom{n}{n/2} \frac{1}{2^n} \approx \sqrt{\frac{2}{\pi n} } \to 0\]

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The Law of Large Numbers says $\displaystyle \mathbb{P}\left[\lim_{n \to \infty} \frac{\# \text{heads} }{n} = \frac{1}{2}\right]=1$.

In other words, the observed fraction of heads always tends to 1/2. These odds can be $1/2 - \epsilon \text{ or }1/2 + \epsilon$.