# \$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.