$a_{n+1} = sqrt{2 + a_n}$ 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:
$$lim_{ntoinfty}a_{n+1}=lim_{ntoinfty}sqrt{2+a_n}$$
Given two functions $f$ and $g$, as long as $f$ is continuous and $lim_{xto x_0}g(x)$ exists, then
$$lim_{xto x_0}f(g(x))=fleft(lim_{xto x_0}g(x)right)$$
This can be shown fairly routinely with the $epsilon ,delta$ definition of the limit.

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

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