Limit involving binomial coefficients without Stirling's formula

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?

Answer

$$l=\lim_{n\to \infty} \frac{(4n)!(n!)^2}{4^n(2n)!^3}=\lim_{n\to\infty}\frac{n!\cdot (2\cdot 2n)(4n-1)(2\cdot (2n-1))...(2n+1)}{2^n2^n(2n)!\cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=\lim_{n\to \infty}\frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
$$\ln l=\lim_{n\to\infty}\sum_{k=1}^{n}ln\left(\frac{2n+2k-1}{2n+2k}\right)$$ And using the fact that $\displaystyle{\lim_{x\to0} \ln(1+x)=x}$ we get that: $$\ln l=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{-1}{2n+2k}=-\frac{1}{2}\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}$$ Which evaluates as a Riemann sum: $$\ln l=-\frac{1}{2}\int_0^1\frac{1}{1+x}dx=-\frac{1}{2}\ln 2\Rightarrow l=\frac{1}{\sqrt2}$$