Finding the value to which a sequence converges The question is $f_1=\sqrt2 \ \ \ , \ \ f_{n+1}=\sqrt{2f_n}$, I have to show that it converges to 2. The book proceeds like this:
let $\lim f_n=l$.
We have, $f_{n+1}=\sqrt{2f_n} \implies (f_{n+1})^2=2f_n$.
Also, $\lim f_n=l \implies \lim f_{n+1}=l$.   [HOW ?]
Thus, $l^2=2l \implies l\in [0,2]$.   [???]
Can someone please explain these two steps.

See this.  In short, you may express the limit as
$$2^{1/2 + 1/4 + 1/8 + \cdots} = 2^1 = 2$$
There are very interesting generalizations.
