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.

Let $g_n = \ln f_n$.
The recurrence becomes
$g_1 = \ln 2/2$,
$g_{n+1} = c+g_n/2$,
where $c = \ln 2/2$.
Calculating the first few terms,
we get
$g_2 = c+g_1/2$,
$g_3 = c + g_2/2
= c+ (c+g_1/2)/2
= 3c/2+g_1/4
$,
$g_4 = c + g_3/2
= c+ (3c/2+g_1/4)/2
= 7c/4+g_1/8
$,
and
$g_5 = c + g_4/2
= c+ (7c/4+g_1/8)/2
= 15c/8+g_1/16
$.
The pattern now seems clear:
$g_n = (2^{n-1}-1)c/2^{n-2}
+ g_1/2^{n-1}
=c(2-1/2^{n-2})
+ g_1/2^{n-1}
$,
and this is easy to prove by induction
from the recurrence.
In the limit,
as $n \to \infty$,
$g_n \to 2c = 2(\ln 2/2) = \ln 2$,
so $f_n \to e^{\ln 2} = 2$.
