# State why lim Xn exists, and find the limit

## Homework Statement

Let x1 = √2, xn+1 = √(2+xn). Use mathematical induction to show that xn < xn+1. Next show that if xn ≥ 2, then xn-1 ≥ 2 also. How do you conclude from this that xn ≤ 2 for all n? State why lim xn 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
Staff Emeritus
Homework Helper
You have

\begin{align*} x_1 & = \sqrt{2} \\ x_2 & = \sqrt{2+\sqrt{2}} \\ x_3 & = \sqrt{2+\sqrt{2+\sqrt{2}}} \\ x_4 & = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \\ & \vdots \end{align*}

Can you see what relationship the limit must satisfy?

tiny-tim
Homework Helper
Hi Jamin2112! If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. Hi Jamin2112! If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. So, since xn+1 = √(2 + xn) and lim xn+1 = lim xn, 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
Homework Helper
2 = √(2 + 2) … what's wrong with that? HallsofIvy
$$\sqrt{2}= 1.414...$$
$$\sqrt{2+ \sqrt{2}}= \sqrt{3.414..}= 1.84776$$
$$\sqrt{2+ \sqrt{2+ \sqrt{2}}= \sqrt{3.84776}= 1.96157$$
$$\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2}}}= \sqrt{3.96157}= 1.99037$$