$a_{n+1} = \sqrt{2 + a_n}$ Specific Theorem Needed

IF the limit exists, it will be a fixed point of the function $\sqrt{2+x}$, in other words a solution to the equation $$x=\sqrt{2+x}$$ Here's why: $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2+a_n}$$ Given two functions $f$ and $g$, as long as $f$ is continuous and $\lim_{x\to x_0}g(x)$ exists, then $$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0}g(x)\right)$$ This can be shown fairly routinely with the $\epsilon ,\delta$ definition of the limit.

Since $\sqrt{2+x}$ is continuous on its domain, $$\lim_{n\to\infty}\sqrt{2+a_n}=\sqrt{2+\lim_{n\to\infty}a_n}=\lim_{n\to\infty}a_{n+1}$$ Since $\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_n$, let $\lim_{n\to\infty}a_n=x$. The initial statement follows.