Let $a_{n+1}=\sqrt{2+a_n},n=1,2,3....$ and $a_1=3$. Then $\underset{n→∞}{\lim}a_n$ is
Answer & explanation
Correct answer: option 2
$\underset{n→∞}{\lim}a_n=\underset{n→∞}{\lim}a_{n+1}=L$, $∴ \underset{n→∞}{\lim}a_{n+1}=\underset{n→∞}{\lim}\sqrt{2+a_n}$
$L=\sqrt{2+L}⇒L^2=2+L⇒L^2-L-2=0$
$L^2-2L+L-2=0,L=2,L=-1$