Answer:
2
Step-by-step explanation:
[tex]a_1=\sqrt2=2^\frac{1}{2}\\\\a_2=\sqrt{2\sqrt2}}=\bigg(2\cdot2^\frac{1}{2}\bigg)^\frac{1}{2}=\bigg(2^\frac{3}{2}\bigg)^\frac{1}{2}=2^{\frac{3}{2}\cdot\frac{1}{2}}=2^\frac{3}{4}\\\\a_3=\sqrt{2\sqrt{2\sqrt{2}}}=\Bigg(2\bigg(2\cdot2^\frac{1}{2}\bigg)^\frac{1}{2}\Bigg)^\frac{1}{2}=\Bigg(2\bigg(2^\frac{3}{2}\bigg)^\frac{1}{2}\Bigg)^\frac{1}{2}=\bigg(2\cdot2^\frac{3}{4}\bigg)^\frac{1}{2}=\bigg(2^\frac{7}{4}\bigg)^\frac{1}{2}\\\\=2^{\frac{7}{4}\cdot\frac{1}{2}}=2^\frac{7}{8}\\\vdots\\\\a_n=2^{\frac{2^n-1}{2^n}}[/tex]
[tex]\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}2^{\frac{2^n-1}{2^n}}=2^{\lim\limits_{n\to\infty}\frac{2^n-1}{2^n}}\qquad(*)}\\\\\lim\limits_{n\to\infty}\dfrac{2^n-1}{2^n}=\lim\limits_{n\to\infty}\bigg(\dfrac{2^n}{2^n}-\dfrac{1}{2^n}\bigg)=\lim\limits_{n\to\infty}\bigg(1-\dfrac{1}{2^n}\bigg)\\\\=\lim\limits_{n\to\infty}1-\lim\limits_{n\to\infty}\dfrac{1}{2^n}=1-0=1\\\\(*)\qquad\lim\limits_{n\to\infty}2^{\frac{2^n-1}{2^n}}=2^1=2[/tex]
