User:J D Bowen/For Julia

Suppose $$x_n\to a < 4 \ $$. Then $$\forall \epsilon>0, \exists N: a-x_n < \epsilon \ \forall n>N \ $$. Set $$4-a=b \ $$ and note

x_{n+1}=\sqrt{3x_n+4} >\sqrt{3(a-\epsilon)+4}=\sqrt{3(4-b-\epsilon)+4} = \sqrt{16-3(b+\epsilon)}