(1)

\begin{align*} \lim_{n\rightarrow\infty}\sqrt{2n}\frac{C(2n,n)}{4^{n}} & =\lim_{n\rightarrow\infty}\sqrt{2n}\frac{\Gamma(2n+1)}{4^{n}\Gamma^{2}(n+1)}\\ & =\lim_{n\rightarrow\infty}\sqrt{2n}\frac{4^{n}\Gamma\left(n+\frac{1}{2}\right)\Gamma(n+1)}{\sqrt{\pi}4^{n}\Gamma^{2}(n+1)}\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}\sqrt{n+\frac{1}{2}}\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{3}{2}\right)}}\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\sqrt{\frac{\sqrt{2n}\sqrt{2n+1}}{2n+1}}\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\right)^{-\frac{1}{4}}\\ & =\sqrt{\frac{2}{\pi}} \end{align*}

(1)-2

\begin{align*} \lim_{n\rightarrow\infty}\sqrt{2n}\frac{C(2n,n)}{4^{n}} & =\frac{2}{\pi}\lim_{n\rightarrow\infty}\sqrt{2n}\int_{0}^{\frac{\pi}{2}}\sin^{2n}\theta d\theta\\ & =\frac{2}{\pi}\sqrt{\frac{\pi}{2}}\\ & =\sqrt{\frac{2}{\pi}} \end{align*}

(2)

\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}4^{n}B(n,n) & =\lim_{n\rightarrow\infty}\sqrt{n}4^{n}\frac{\Gamma(n)\Gamma(n)}{\Gamma(2n)}\\ & =2\sqrt{\pi}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\\ & =2\sqrt{\pi} \end{align*}

(3)

\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\left(1-t^{2}\right)^{n}dt & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}4^{n+1}B(n+1,n+1)\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\sqrt{n+1}}\sqrt{n+1}4^{n+1}B(n+1,n+1)\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}4^{n}B(n,n)\\ & =\sqrt{\pi} \end{align*}