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

• A. $-1$
• B. $2$
• C. $\sqrt{5}$
• D. $3$

Answer 1

Answer: (B)

$2$

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