Answer:
3
Step-by-step explanation:
hello,
let's define the following sequence
[tex]a_0=\sqrt{6}\\a_{n+1}=\sqrt{6+a_n}[/tex]
the number we are looking for is the limit of the sequence
we can prove that the sequence is increasing and
[tex]a_n <= 3[/tex] by induction
(by noticing that if [tex]a_n <=3[/tex] then [tex]a_{n+1}=\sqrt{6+a_n} <=\sqrt{6+3} =3[/tex] )
so the limit of the sequence exists, let's note it l we can write from
[tex]a_{n+1}=\sqrt{6+a_n}[/tex]
that
[tex]l=\sqrt{6+l}[/tex]
so
[tex]l^2=6+l <=> l^2-l-6=0 <=> l^2-3l+2l-6=0\\<=> l(l-3)+2(l-3)=0<=>(l+2)(l-3)=0\\\\<=>l=-2 \ \ or \ \ l = 3[/tex]
as
[tex]a_n >= 0[/tex] it comes l = 3
hope this helps
