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

**Lambin** · December 21st 2012, 06:34 PM

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?

**Prove It** · December 21st 2012, 09:05 PM

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

**hollywood** · December 22nd 2012, 09:43 AM

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

Originally Posted by Lambin

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

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

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

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?

In the last step you substituted 0 for $x_n-2$ . But it is not true that $x_n-2=0$ 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

**x3bnm** · December 22nd 2012, 06:21 PM

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

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

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

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

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

**Lambin** · December 23rd 2012, 12:37 PM

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

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

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

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

LinkBack

Thread Tools

Display

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

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

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

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

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

Similar Math Help Forum Discussions

What is the limit of x^(1/x) as x approaches infinity?

Limit as X approaches infinity

Limit as x approaches infinity

limit as x approaches -infinity?

the limit as x approaches infinity

Search Tags