# Thread: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

1. ## Evaluation of a limit as n approaches infinity using L'Hopital's Rule

PROBLEM:

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

x_0 = 1

and,

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

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

ATTEMPT:

So far,

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

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

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

L'Hopital's rule:

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

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

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

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?

2. ## Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

To use LaTeX on this forum, you need to put the code in tex tags, which look like [ tex ] [ / tex ] without the spaces.

3. ## Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

I stripped out the mimetex stuff and was able to make the question readable:

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?
In the last step you substituted 0 for $\displaystyle x_n-2$. But it is not true that $\displaystyle x_n-2=0$ for all n, so you can't make that substitution.

I put the problem into Excel, and it looks like the limit should be about -1.097, but I can't see how to get to that result algebraically. Can someone else help?

- Hollywood

4. ## Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

Sorry I tested on my maple the answer is $\displaystyle 0$:

\displaystyle \begin{align*}\therefore \lim_{n\to \infty} 4^n (x_n - 2)=& \lim_{n\to \infty} 4^n \left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ \cdots +1}}}}} - 2\right) \\ =& 0\end{align*}

Is there a way to show that this limit $\displaystyle 0$ can be found mathematically?

How can one find limit of $\displaystyle \lim_{n\to \infty } 4^{n}(x_{n} - 2) = \lim_{n\to \infty} 4^n \left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ \cdots +1}}}}} - 2\right)$

using math(without the help of computer) where $\displaystyle x_{n} = \sqrt{2 + x_{n-1}} \text{ and } x_0 = \sqrt{(2+1)}$?

5. ## Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

Thank you all for trying to help.

And sorry for the trouble of reading this post; I just now started learning LaTeX.

Here's the solution:

Put $\displaystyle x_n=2y_n$ to get $\displaystyle y$_{n+1}$=\sqrt{\frac{1+y_n}{2}}$ and $\displaystyle y_0=\frac{1}{2}=cos(\frac{\pi}{3})$

Therefore, $\displaystyle y_n=cos(\frac{\pi}{3*2^n})\implies x_n=2cos\frac{\pi}{3*2^n}$

Thus, $\displaystyle z$_{n}=4$^{n}(x_n-2)=\frac{-2(1-cos\frac{\pi}{3*2^n})}{(\frac{\pi}{3*2^n})^2}\frac {\pi}{3*2^n}^2}(4^n)$

Hence,$\displaystyle$\lim_{n\to\infty}z_n=-2*\frac{1}{2}*\frac{\pi^2}{9}=\frac{-\pi^2}{9} \approx -1.097