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

[kathrynag]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[Mark44]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

No, you can't use this |x^{1/2}-\infty|. The definition for the limit of an unbounded function doesn't use infinity or epsilon. Do you know what that definition is?

 

[kathrynag]

Well, I was supposed to supply the definition of lim x_{n}=\infty.

I guess I wasn't so sure on how to do that.

 

[Mark44]

No, you're supposed to use the definition of \lim_{n \to \infty} x_n = \infty. That's different.

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

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.

 

[kathrynag]

Sorry, I was just using what my book stated and the book didn't include that extra part.

 

[Mark44]

Do you still have a question? I can't tell.

 

[kathrynag]

I guess my question is how to use the definition to prove the limit of my original sequence?

 

[Mark44]

You work backward. Assuming for the moment that n^1/2 > M, can you find some other number N so that when n > N, then n^1/2 > M? That's basically what I said in post #4.