Is n over itself, over itself ... recursively equal to one, or square-root of n?

Question

Suppose that $x = \frac{n}{x} $ where n a constant real number and we wish to solve for x. Solving analytically, $x^2 = n \rightarrow x = \sqrt{n}$. But, suppose we plugged in the value of x recursively so that $x = \frac{n}{\frac{n}{\frac{n}{...}}}$. Does this mean that $x = \frac{n}{\frac{n}{\frac{n}{...}}} = \sqrt{n}$? Intuitivly, I would expect it to be 1.

Answer 1

Usually we define infinite nesting as the $k\to\infty$ limit of the result with $k$ items. In this case the sequence is $x_0=n,\,x_{k+1}=n/x_k$, a sequence that alternates bewteen $n$ and $1$, and for $n\ne1$ doesn't converge.