# Thread: Squeeze Rule? R2 limit

1. But if we considered only integers, then would the squeeze theorem argument work?

2. Originally Posted by shilz222
But if we considered only integers, then would the squeeze theorem argument work?
no, because the inequality you have is, in fact, not true. it is only true for some x and y, and it is not true for the x's and y's that are close to zero, which is the point we are considering. if we are dealing with integers, the closest values of x and y would be +/-1 and it is not true for those

3. Ok, so you cant use the squeeze rule at all for this problem?

4. Originally Posted by shilz222
Ok, so you cant use the squeeze rule at all for this problem?
no, i do not see a way to use it effectively. since we know the answer before hand, we know that if the squeeze theorem would work, we would have to find a way to bound it above and below by 2. there is no way we would be able to guess 2 right off the bat. rationalizing was you best bet here, and will usually be your best bet for limits off this type, so mark it well

5. Isnt the inequality true for $-1$ and $1$ (the closest integers to $0$)? So would this limit be $0$ in $\mathbb{Z}^{2}$?

6. Originally Posted by shilz222
Isnt the inequality true for $-1$ and $1$ (the closest integers to $0$)? So would this limit be $0$ in $\mathbb{Z}^{2}$?
I've been reading over this post and since Jhevon isn't around at the moment I'll answer this.

I think Jhevon has done an excellent job at explaining why you can't take a limit under the set of integers, of any dimensionality. So the answer is a resounding no!
because you simply can't set up the limit and make any sense of it.

The set of integers is not continous, unless you wish to examine other topologies. (And I'm not sure you can get away with that. You'd have to ask Plato if that's even possible.) I think it is safe to presume that any given limit is being done on the set of real numbers, unless otherwise specified.

-Dan

7. Is there some weird combination of $\mathbb{R}$ and $\mathbb{Z}$(thus continuous) that could possibly make the limit $0$?

8. Originally Posted by shilz222
Is there some weird combination of $\mathbb{R}$ and $\mathbb{Z}$(thus continuous) that could possibly make the limit $0$?
If I am not mistaken it is possible to make the integers continuous in the discrete topology. (Though I'm not going to stand behind that statement. This conclusion is based on a brief web search and my spotty memory. I was never good with continuity arguments!)

However even in this case the limit is 2, not 0. I can't think of a way to force this expression to be anything other than 2 as long as we use a subset or extension of the reals. And if we don't use such a set, the comparison of this limit to the limit under the reals is meaningless.

-Dan

9. If we make the integers continuous in a discrete topology, then why doesn't my argument hold (squeeze theorem)? Sorry for beating down a dead horse.

10. Originally Posted by shilz222
If we make the integers continuous in a discrete topology, then why doesn't my argument hold (squeeze theorem)? Sorry for beating down a dead horse.
I'm not saying I'm sure about my own result, I just can't see any logical way that it would be different and still make logical sense.

To do the job properly you'd have to study the discrete topology and see if the squeeze theorem still applies. And I'm nowhere near good enough in topology to even consider trying.

-Dan

11. Originally Posted by shilz222
Isnt the inequality true for $-1$ and $1$ (the closest integers to $0$)? So would this limit be $0$ in $\mathbb{Z}^{2}$?
no, the inequality is not true for +/- 1.

Originally Posted by shilz222
$0 \leq \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} \leq x^{2} + y^{2}$. So wouldn't the limit be $0$?
whether we plug in +1 or -1 is immaterial, since we would square it and get 1. so let's plug in 1 without loss off generality

plug in x = y = 1

the LHS gives: $\frac 2{\sqrt{3} - 1} \approx 2.7321$

the RHS gives: $1^2 + 1^2 = 2$

obviously $2.7321 \not \le 2$

so the inequality is false here.

again, without loss of generality, we can make x = 0, y = +/- 1

the LHS gives: $\frac 1{\sqrt{2} - 1} \approx 2.414$

the RHS gives: $0 + 1^2 = 1$

again, it is obvious that $2.414 \not \le 1$

so again, your inequality is false. i have just covered all possibilities close to zero for the integers, not very interesting huh? and i was pushing it with the last one, since i let x = 0 and not just be close to zero

in all cases, you inequality does not hold

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