State why lim Xn exists, and find the limit

In summary: I would also like to point out, that if the limit exists, then all terms in the sequence must approach the limit. This is because the recursive definition of xn+1, depends on xn. Therefore, if xn+1 approaches the limit, then xn must also approach the limit. This is why xn < xn+1. In summary, using mathematical induction, it can be shown that xn < xn+1 for all n. Additionally, with xn ≥ 2, it can be shown that xn-1 ≥ 2 as well. This implies that xn ≤ 2 for all n. The limit of the sequence is found to be 2, as shown by substituting the limit into the recursive definition of xn+1
  • #1
Jamin2112
986
12

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?
 
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  • #2
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
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
tiny-tim said:
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
2 = √(2 + 2) … what's wrong with that? :smile:
 
  • #6
[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?
 

1. What does it mean for a limit to exist?

A limit exists when the values of a function approach a certain value as the input approaches a specific value, without ever reaching that specific value.

2. How can you prove that a limit exists?

To prove that a limit exists, you must show that the values of the function get closer and closer to a single value as the input gets closer to a specific value. This can be done through various methods such as using the formal definition of a limit, graphing the function, or using algebraic techniques.

3. What is the formal definition of a limit?

The formal definition of a limit states that the limit of a function at a specific point is the value that the function approaches as the input approaches that specific point. This can be written as: lim x->a f(x) = L, where L is the limit and a is the point the input is approaching.

4. Can a limit exist even if the function is not defined at that point?

Yes, a limit can exist even if the function is not defined at that point. This is because the limit is based on the behavior of the function as the input approaches a specific value, not the actual value at that point.

5. What are some common techniques for finding limits?

Some common techniques for finding limits include using algebraic manipulation, factoring, and using limit laws (such as the sum, difference, and product laws). Additionally, graphing the function or using the formal definition of a limit can also be helpful in finding limits.

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