Q:

Finding a limit of a sequence if nth term is not given

I have that $x_1 = \sqrt{2}$ and $x_n = \sqrt{ 2 \sqrt{x_{n-1} } } $. After working out details, I obtain that 
$$ x_n = \sqrt{ 2 \cdot 2^{ 1/2^n} } = \sqrt{ 2^{ (2^n+1)/2^n} } = 2^{(2^n + 1)/2^{n+1}} \to \sqrt{2} $$
Is this the correct limit?

A:

Assume convergence first, hence for $\lim_{n\to \infty} x_n=\lim_{n\to \infty} x_{n-1}=x$.
We get $x=(2(x)^{1/2})^{1/2}=2^{1/2}x^{1/4}$. Take the forth power to get the following.
$$x^4=4x$$
$$x^3=4$$
$$x=\sqrt[3]{4}$$

