State why lim Xn exists, and find the limit

[Jamin2112]

Homework Statement

Let x₁ = √2, x_n+1 = √(2+x_n). Use mathematical induction to show that x_n < x_n+1. Next show that if x_n ≥ 2, then x_n-1 ≥ 2 also. How do you conclude from this that x_n ≤ 2 for all n? State why lim x_n exists, and find the limit.

Homework Equations

Upper bounds, limits, etc.

The Attempt at a Solution

I've done everything up to the point of "find the limit." Help?

 

[vela]

You have

\begin{align*}<br /> x_1 &amp; = \sqrt{2} \\<br /> x_2 &amp; = \sqrt{2+\sqrt{2}} \\<br /> x_3 &amp; = \sqrt{2+\sqrt{2+\sqrt{2}}} \\<br /> x_4 &amp; = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \\<br /> &amp; \vdots<br /> \end{align*}

Can you see what relationship the limit must satisfy?

 

[tiny-tim]

Hi Jamin2112!

If the limit exists, and is x, then you can substitute x for x_n and x_n+1 in the same equation.

 

[Jamin2112]

Hi Jamin2112!

If the limit exists, and is x, then you can substitute x for x_n and x_n+1 in the same equation.

So, since x_n+1 = √(2 + x_n) and lim x_n+1 = lim x_n, we can denote the limit A, and solve for A as such:

A = √(2+A)
A² = 2 + A
A² - A - 2 = 0
(A-2)(A+1) = 0
A = 2 , -1

Therefore A = 2.

That seems wrong.

 

[tiny-tim]

2 = √(2 + 2) … what's wrong with that?

 

[HallsofIvy]

\sqrt{2}= 1.414...
\sqrt{2+ \sqrt{2}}= \sqrt{3.414..}= 1.84776
\sqrt{2+ \sqrt{2+ \sqrt{2}}= \sqrt{3.84776}= 1.96157
\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2}}}= \sqrt{3.96157}= 1.99037

Why would "2" not seem right?