State why lim Xn exists, and find the limit

  • Thread starter Jamin2112
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  • #1
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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?
 

Answers and Replies

  • #2
vela
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You have

[tex]\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*}[/tex]

Can you see what relationship the limit must satisfy?
 
  • #3
tiny-tim
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Hi Jamin2112! :smile:

If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. :wink:
 
  • #4
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Hi Jamin2112! :smile:

If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. :wink:
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.
 
  • #5
tiny-tim
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2 = √(2 + 2) … what's wrong with that? :smile:
 
  • #6
HallsofIvy
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[tex]\sqrt{2}= 1.414...[/tex]
[tex]\sqrt{2+ \sqrt{2}}= \sqrt{3.414..}= 1.84776[/tex]
[tex]\sqrt{2+ \sqrt{2+ \sqrt{2}}= \sqrt{3.84776}= 1.96157[/tex]
[tex]\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2}}}= \sqrt{3.96157}= 1.99037[/tex]

Why would "2" not seem right?
 

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