Question

If $x_1=3$ and $x_{n+1}=\sqrt{2+x_n},n\ge 1$, then $\underset{n\to \infty }{lim}{x}_n$ is​​​​​​​

Answer 1

$x_{n+1}=\sqrt{2+x_n}$
$\Rightarrow \underset{n\to \infty }{lim}{x}_{n+1}=\sqrt{2+\underset{n\to \infty }{lim}{x}_n}$
$t=\sqrt{2+t}(\because \underset{n\to \infty }{lim}{x}_{n+1}=\underset{n\to \infty }{lim}{x}_n=t)$
$\Rightarrow t^2-t-2=0$
$\Rightarrow (t-2)(t+1)=0\Rightarrow t=2(\because t>0)$