Let S=√(2+√(2+√(2+√(2))))...
Then S^2=2+(√(2+√(2+√(2+√(2))))...)
or S^2=2+S
S^2-S-2=0
(S-2)(S+1)=0
so S=2
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for the general "x"
S=√(x+√(x+√(x+√(x))))...
S^2=x+S
S^2-S-x=0
no easy factoring this time, so use:
(-b±(b^2-4ac)^.5)/2a
and use + rather than -
so S=(1+(1+4x)^.5)/2
in the limit, as x goes to infinity,
S approaches √x