How to Show lim n^{1/2}=\infty using the Definition of lim x_{n}=\infty?

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    Infinity Limit
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Homework Help Overview

The discussion revolves around demonstrating that the limit of \( n^{1/2} \) approaches infinity using the formal definition of limits for sequences. Participants are exploring the nuances of applying the definition of limits to an unbounded function.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the correct definition of limits as \( n \) approaches infinity and how to apply it to the sequence \( n^{1/2} \). Questions arise regarding the proper formulation of the limit definition and the interpretation of the variables involved.

Discussion Status

The discussion is ongoing, with participants providing insights into the formal definition of limits and questioning the original poster's understanding of the necessary steps. Some guidance has been offered regarding the approach to take, but no consensus has been reached on how to proceed with the proof.

Contextual Notes

There appears to be some confusion regarding the definitions provided in the original poster's textbook, which may not fully align with the expectations for this problem. Participants are also addressing the need for clarity on the definitions used in the context of limits approaching infinity.

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Homework Statement


Show lim [tex]n^{1/2}[/tex]=[tex]\infty[/tex] using the definition of lim [tex]x_{n}[/tex]=[tex]\infty[/tex].



Homework Equations





The Attempt at a Solution


We want to show [tex]\left|x^{1/2}-\infty\right|[/tex],[tex]\epsilon[/tex]. I get this far and i hit a blank.
 
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kathrynag said:

Homework Statement


Show lim [tex]n^{1/2}[/tex]=[tex]\infty[/tex] using the definition of lim [tex]x_{n}[/tex]=[tex]\infty[/tex].



Homework Equations





The Attempt at a Solution


We want to show [tex]\left|x^{1/2}-\infty\right|[/tex],[tex]\epsilon[/tex]. I get this far and i hit a blank.
No, you can't use this [tex]|x^{1/2}-\infty|[/tex]. The definition for the limit of an unbounded function doesn't use infinity or epsilon. Do you know what that definition is?
 
Well, I was supposed to supply the definition of lim [tex]x_{n}[/tex]=[tex]\infty[/tex].

I guess I wasn't so sure on how to do that.
 
No, you're supposed to use the definition of [tex]\lim_{n \to \infty} x_n = \infty[/tex]. That's different.

[tex]\lim_{x \to \infty} f(x) = \infty[/tex] is defined this way:
[tex]\forall M \exists N > 0 \ni \forall x > N, f(x) > M[/tex]

In other words, no matter how large an M someone chooses, there is some number N so that if x > N, then f(x) > M. In other, other words, if you want to get a larger value for f(x), take a larger value for x.

The definition is similar for your sequence.
 
Sorry, I was just using what my book stated and the book didn't include that extra part.
 
Do you still have a question? I can't tell.
 
I guess my question is how to use the definition to prove the limit of my original sequence?
 
You work backward. Assuming for the moment that n1/2 > M, can you find some other number N so that when n > N, then n1/2 > M? That's basically what I said in post #4.
 

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