Originally Posted by

**Lambin** **PROBLEM:**

$\displaystyle (x_{n})$-series of numbers where:

$\displaystyle x_0 = 1$

and,

$\displaystyle x_{n+1} = sqrt(2+x_{n})$

Find: $\displaystyle \lim_{n\rightarrow \infty } 4^{n}(x_{n} - 2) = ?$

**ATTEMPT:**

So far,

$\displaystyle y\,=\,\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\,.\,.\,.\sq rt{3}}}}}$

$\displaystyle y^2\,=\,2\,+\,y$

$\displaystyle (y\,-\,2)(y\,+\,1)\,=\,0\,\Rightarrow\,y\,=\,2$

L'Hopital's rule:

$\displaystyle \lim_{n\to\infty}\,\frac{x_n\,-\,2}{\frac{1}{4^n}}$

$\displaystyle \lim_{n\to\infty}\,\frac{0}{\frac{-4^n\ln(4)}{4^{2n}}}\,=\,0$

$\displaystyle \lim_{n\to\infty } 4^{n}(x_{n}\,-\,2) = 0$

My application of L'Hopital's rule is incorrect. It's an indeterminate form, but not resolved in the way I indicated. Any ideas on how I could resolve the problem?