Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$$
My attempt,
Let $y_n = \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}$, we see that $ \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6\cdot ...\cdot 2n} > \frac{1 \cdot 2 \cdot 2 \cdot ... \cdot 2}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n} = \frac{1}{1 \cdot 2 \cdots .. \cdot (n-1)\cdot2n} = x_n$
$$\lim_{n \to \infty}{x_n} = 0 $$
Now I just need to find a sequnce $z_n>y_n$, so, $\lim_{n \to \infty} z_n = 0$. 

Hint multiply both numerator,denominator by $2.4.6....2n$ so you get $$\frac{(2n)!}{4^n(n!)^2}$$ where i get $4^n$ by taking $2$ common so we can write it as $$\frac{{2n\choose n}}{4^n}$$ thus we know $4^n\geq {2n\choose n}$. So limit is $0$.
