# Squeeze Rule? R2 limit

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 18th 2007, 10:13 AM
shilz222
Squeeze Rule? R2 limit
Can you use the squeeze rule for the following: $\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1}$ rather than multiplying by the conjugate?

I did the following: $\displaystyle 0 \leq \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} \leq x^{2} + y^{2}$. So wouldn't the limit be $\displaystyle 0$?
• Oct 18th 2007, 10:39 AM
Plato
Quote:

Originally Posted by shilz222
$\displaystyle 0 \leq \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} \leq x^{2} + y^{2}$.

Try that inequality for x=0.1 & y=0.2.
• Oct 18th 2007, 01:47 PM
shilz222
What a dumb mistake. Aargh.
• Oct 18th 2007, 01:57 PM
Plato
Can you use polar substitution?
$\displaystyle r^2 = x^2 + y^2 ,\;x = r\cos (t),\;y = r\sin (t)$.

You will get a function of $\displaystyle r$ alone and then let $\displaystyle r \to 0$
• Oct 18th 2007, 02:11 PM
shilz222
yes it would be $\displaystyle \frac{r^{2}}{\sqrt{r^2+1}-1}$ so now you can use L'Hopitals rule.
• Oct 18th 2007, 02:16 PM
Plato
Very good. Is the limit 2?
• Oct 18th 2007, 02:21 PM
shilz222
yes. the limit is 2. I made a mistake on that one on a test.
• Oct 18th 2007, 02:38 PM
ThePerfectHacker
Just rationalize!
• Oct 18th 2007, 02:41 PM
shilz222
Yes I missed that. I knew I should have done that. But alas, I was under a time constraint.
• Oct 20th 2007, 11:33 PM
shilz222
Can you say the limit is 0 if we restrict $\displaystyle x \in \mathbb{Z}$? Or is this not allowed?
• Oct 20th 2007, 11:39 PM
Jhevon
Quote:

Originally Posted by shilz222
Can you say the limit is 0 if we restrict $\displaystyle x \in \mathbb{Z}$? Or is this not allowed?

it's not allowed. i think the question was meant for real numbers, otherwise, we could not make quantities arbitrarily close to each other and the limit would make no sense then
• Oct 21st 2007, 12:28 AM
shilz222
But we have Real Analysis and we have Complex Analysis. So why cant we consider the limit in $\displaystyle \mathbb{Z}^{n}$? (i.e. Integer Analysis)? Couldn't we also consider this same limit in $\displaystyle \mathbb{C}^{n}$? It wouldnt make sense in $\displaystyle \mathbb{R}^{n}$ but would it make sense in $\displaystyle \mathbb{Z}^{n}$?
• Oct 21st 2007, 12:31 AM
Jhevon
Quote:

Originally Posted by shilz222
But we have Real Analysis and we have complex analysis. So why cant we consider the limit in $\displaystyle \mathbb{Z}^{n}$? (i.e. Integer Analysis)? Couldn't we also consider this same limit in $\displaystyle \mathbb{C}^{n}$?

$\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{C}$ are complete sets (the completeness axiom applies). this is a property that makes it possible to evaluate limits, because we need to make things as close as possible. this is not possible with the integers, since there are gaps 1 unit in length between integers. finding limits under those conditions make no sense really, since we can't make things arbitrarily close (recall the epsilon-delta definition of the limit of a function)
• Oct 21st 2007, 12:35 AM
shilz222
But the definition of the limit (epsilon-delta) is based on $\displaystyle \mathbb {R}^{n}$? So we cant really use a definition of a limit that is defined in $\displaystyle \mathbb{R}^{n}$ in $\displaystyle \mathbb{Z}^{n}$ right? So basically when you take a limit in $\displaystyle \mathbb{Z}^{n}$ the closest you can be is 1 unit away. So cant you restrict the definition of a limit? You are still taking a limit right?

Actually I think you can. $\displaystyle \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset$ $\displaystyle \mathbb{R}$ by the Dedekind cuts. So the definition that applies in $\displaystyle \mathbb{R}^{n}$ will also apply to $\displaystyle \mathbb{Z}^{n}$ but not vice-versa.
• Oct 21st 2007, 12:49 AM
Jhevon
Quote:

Originally Posted by shilz222
But the definition of the limit (epsilon-delta) is based on $\displaystyle \mathbb {R}^{n}$? So we cant really use a definition of a limit that is defined in $\displaystyle \mathbb{R}^{n}$ in $\displaystyle \mathbb{Z}^{n}$ right? So basically when you take a limit in $\displaystyle \mathbb{Z}^{n}$ the closest you can be is 1 unit away. So cant you restrict the definition of a limit? You are still taking a limit right?

Actually I think you can. $\displaystyle \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset$ $\displaystyle \mathbb{R}$ by the Dedekind cuts. So the definition that applies in $\displaystyle \mathbb{R}^{n}$ will also apply to $\displaystyle \mathbb{Z}^{n}$ but not vice-versa.

we can take limits in $\displaystyle \mathbb{C}$ as well. even if you could come up with a construction for limits in the integers, it would be essentially meaningless. since it would give rise to things like continuity at a point where there is actually a gap. it would simply make no sense, and won't be very interesting. in any case, if you considered only integers here, the answer would still be 2, so why even bother with it?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last