12. limits — ગણિત ધોરણ 11 સાયન્સ — Question

${x_2} = \sqrt {2 + {x_1}} = \sqrt {2 + 3} = \sqrt 5 $, $\,{x_3} = \sqrt {2 + {x_2}} = \sqrt {2 + \sqrt 5 } $

$\therefore \,\,\,{x_1} > {x_2} > {x_3}$

It can be easily shown by mathematical induction that the sequence ${x_1},\,\,{x_2},........{x_n},....$ is a monotonically decreasing sequence bounded below by $2$.

So it is convergent. Let $\lim {x_n} = x.$ Then

${x_{n + 1}} = \sqrt {2 + {x_n}} \, \Rightarrow \,\,\lim {x_{n + 1}} = \sqrt {2 + \lim {x_n}} $$ \Rightarrow \,x = \sqrt {2 + x} $

$ \Rightarrow \,\,{x^2} - x - 2 = 0\,\, \Rightarrow \,\,(x - 2)\,(x + 1) = 0\, \Rightarrow \,x = 2$