If $b_{n+1}=\sqrt{2+b_{n}}$ for $n=1,2,3,\cdots\cdots$. Then $\lim_{n\rightarrow\infty}b_{n}$ is , given $b_{1}=3$
Try: let $b_{n}=2\cos(x_{n})$. Then put into $b_{n+1}=\sqrt{2+b_{n}}$
We have $$2\cos(x_{n+1})=2\cos\bigg(\frac{x_{n}}{2}\bigg)$$
Could some help me how to calculat $\lim_{n\rightarrow\infty}b_{n}$
Thanks in advanced
For the limit $L = \lim_{n\to\infty} b_n$ we have $L=\sqrt{2+L}$ by continuity of $f(x)=\sqrt{2+x}$. Can you now solve the equation to find out the single possibility for $L$?
Method 1:
Your substitution is a good one.
Note that $b_1=3\ge0$. Let $\displaystyle x_1=\arccos\left(\frac{b_1}{2}\right)$. Then $\displaystyle x_1\in\left[0,\frac{\pi}{2}\right]$.
For $n>1$, let $\displaystyle x_{n+1}=\frac{x_n}{2}$.
Since $\displaystyle x_{n}=\frac{x_1}{2^{n-1}}$, $\displaystyle \lim_{n\to\infty}x_n=0$.
Note that $b_1=2\cos(x_1)$. If $b_k=2\cos(x_k)$. Then
$$b_{k+1}=\sqrt{2+b_k}=2\sqrt{\cos^2\left(\frac{x_k}{2}\right)}=2\cos\left(\frac{x_k}{2}\right)=2\cos(x_{k+1})$$
By mathematical induction, $b_n=2\cos(x_n)$ for all $n\ge 1$.
$$\lim_{n\to\infty}b_n=2\cos(\lim_{n\to\infty}x_n)=2\cos(0)=2$$
Method 2:
$b\ge 2$ $\implies$ $b^2\ge2 b\ge b+2\ge 4$ $\implies$ $b\ge \sqrt{b+2}\ge 2$.
Since $b_1=3\ge 2$, then $\{b_n\}$ is decreasing and bounded below by $2$. It is convergent.
Let $\displaystyle \lim_{n\to \infty}b_n=\ell$. Then $\ell>0$ and
\begin{align*} \lim_{n\to \infty}b_{n+1}&=\lim_{n\to \infty}\sqrt{2+b_n}\\ \lim_{n\to \infty}b_{n+1}&=\sqrt{2+\lim_{n\to \infty}b_n}\\ \ell&=\sqrt{2+\ell}\\ \ell^2-\ell-2&=0\\ \ell&=2 \end{align*}
