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Let $a_{n+1}=\sqrt{2+a_n},n=1,2,3....$ and $a_1=3$. Then $\underset{n→∞}{\lim}a_n$ is |
1 2 $\sqrt{5}$ 3 |
2 |
$\underset{n→∞}{\lim}a_n=\underset{n→∞}{\lim}a_{n+1}=L$, $∴ \underset{n→∞}{\lim}a_{n+1}=\underset{n→∞}{\lim}\sqrt{2+a_n}$ $L=\sqrt{2+L}⇒L^2=2+L⇒L^2-L-2=0$ $L^2-2L+L-2=0,L=2,L=-1$ |
