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

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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.
 
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.