Tìm các giới hạn sau
$I_1=\lim_{x\rightarrow +\infty}\left(2^n.\sqrt{2-\underset{n-1}{\underbrace{\sqrt{2+\sqrt{2+...+\sqrt{2}}}}}}\right)$
$I_2=\lim_{x\rightarrow+\infty}\frac{\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}....\underset{n}{\underbrace{\sqrt{2+\sqrt{2+...+\sqrt{2}}}}}}{2^n}$
$I_1=\lim_{x\rightarrow +\infty}\left(2^n.\sqrt{2-$2.cos\frac{\pi }{2^{^{n}}}$ = $\lim_{x\rightarrow +\infty}\left(2^n.\sqrt{$2.(1-cos\frac{\pi }{2^{^{n}}})$ = $\lim_{x\rightarrow +\infty}\left(2^n.\sqrt{$2.2$sin^{2}\frac{\pi }{2^{n+1}}$ = $\lim_{x\rightarrow +\infty}\left(2^{n+1}.$sin\frac{\pi }{2^{n+1}}$ = 1