Differentiation Question 129
Question: If $ x_1=3 $ and $ {x _{n+1}}=\sqrt{2+x _{n},}n\ge 1, $ 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:
[b] $ {x _{n+1}}=\sqrt{2+x _{n}}\Rightarrow \lim {x _{n+1}}=\sqrt{2+\lim x _{n}} $
$ \Rightarrow t=\sqrt{2+t} $
$ [\because \lim {x _{n+1}}=limx _{n}] $
$ \Rightarrow t^{2}-t-2=0\Rightarrow (t-2)(t+1)=0\Rightarrow t=2. $
BETA
