Functions Question 617
Question: If $ x_1=3 $ and $ x>0 $ then $ \underset{n\to \infty }{\mathop{\lim }},x_{n} $ is equal to
Options:
A) -1
B) 2
C) $ \sqrt{5} $
D) 3
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ x_1=3,{x_{n+1}}=\sqrt{2+x_{n}} $
$ 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 $
$ (\because x_{n}>0,\forall ,n;\therefore x>0) $
BETA
