Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle  \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ 

Note that
$$\frac{{1.3{\cdots}(2n - 1)}}{{2.4{\cdots}(2n)}} = \frac{{\frac{{(2n - 1)!}}{{2.4{\cdots}(2n - 2)}}}}{{{2^n}n!}} = \frac{{\frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}}}}{{{2^n}n!}} = \frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}} \cdot \frac{1}{{{2^n}n!}} \to 0$$
using Stirling's approximation because $n$ is "large" here.
