Answer
$\lim\limits_{n \to \infty}a_{n} = 2$
Work Step by Step
$a_{n} =$ {$\sqrt 2, \sqrt (2\sqrt 2), \sqrt (2\sqrt (2(\sqrt 2)),...$}
$=$ {$2^{\frac{1}{2}}, 2^{\frac{1}{2}} \times 2^{\frac{1}{4}}, 2^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 2^{\frac{1}{8}}, ...$}
$=$ {$2^{\frac{1}{2}}, 2^{\frac{3}{4}}, 2^{\frac{7}{8}}, ...$}
$a_{n} = 2^{\frac{(2^{n}-1)}{2^{n}}}$
Taking limit as $nā \infty$
$\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} 2^{\frac{(2^{n}-1)}{2^{n}}}$
$= 2^{\lim\limits_{n \to \infty}\frac{(2^{n}-1)}{2^{n}}}$
$\lim\limits_{n \to \infty}\frac{2^{n}-1}{2^{n}}= \lim\limits_{n \to \infty}\frac{1-\frac{1}{2^{n}}}{1}$
$=\frac{1-0}{1}=1$
$\lim\limits_{n \to \infty}a_{n} = 2^{1}$
$\lim\limits_{n \to \infty}a_{n} = 2$
