I have this question from a friend who is taking college admission exam, evaluate: $$\lim_{n\to\infty} \frac{\binom{4n}{2n}}{4^n\binom{2n}{n}}$$ The only way I could do this is by using Stirling's formula:$$ n! \sim \sqrt{2 \pi n} (\frac{n}{e})^n$$ after rewriting as $$\lim_{n\to\infty} \frac{(4n)!(n!)^2}{4^n(2n)!^3}$$ and it simplifies really satisfying to $\frac{1}{\sqrt2}$.
However Stirling's formula is not in the syllabus nor taught in high school, is there an elementary approach to this limit?