Talk:Squeeze theorem

-->I included References, but I am not good with the notation of citation, if anybody could do something about any errors I made it would be appreciated. EulerGamma 02:28, 1 September 2006 (UTC)

"Kathy Theorem"
Since when is this known as the "Kathy Theorem"? A cursory Google doesn't turn up much and I'm inclined to think it's a vanity/joke edit without any references Cxseven (talk) 12:47, 14 June 2015 (UTC)

Italian name

 * In Italian, the squeeze theorem is also known as the two carabineri theorem. The idea is that two carabinieri are holding a prisoner in the middle.

How is this relevant to English-speaking readers? —Frungi 03:56, 26 September 2006 (UTC)


 * Because it is an effective visualisation of what exactly is going on in the situation; use your imagination. Dysprosia 05:02, 26 September 2006 (UTC)


 * I always imagined myself to be in the middle. This did help my comprehension, although not making me feel as comfortable as I would be with the idea of a sandwich. I regret that they did not teach me the theorem with the other name. Still, I keep using the Italian form, in my classes. --PMajer (talk) 12:32, 30 October 2008 (UTC)

- The link for Dr. C. Sean Bohun's paper is broken...--Yfarjoun 21:48, 5 November 2007 (UTC)

Real world uses
Are there any real world uses for this function that can be used to bolster the definition? At first blush this seems like a merely academic theorem/painfully obvious notion. What's the purpose of the squeeze function for applications? —Preceding unsigned comment added by 76.24.24.60 (talk) 13:33, 12 November 2007 (UTC) It's useful for evaluating all sorts of limits For example it can be used to show the derivative of sinx is cosx which is a basic fact that's used implicitly throughout maths and physics. —Preceding unsigned comment added by 86.128.127.8 (talk) 20:25, 18 May 2008 (UTC)
 * It's a useful, even if small, theorem. That the theorem exists and is proven makes writing some things easier - in stead of going through the proof in each individual case, you can show that your situations matches the criteria for this theorem and move directly to the result. As indicated in the preceding comment, it is useful for some simple proofs especially. 81.147.91.113 (talk) 17:51, 19 April 2009 (UTC)

"Proof"
The second section entitled "Proof" is entirely incorrect. It claims to show that f(x) = L, the limit of the 'squeezing' functions. That's not true. The limit of f(x) at that point is L. I have therefore removed this section as it is wrong; it is perhaps worthwhile that someone fix and reinstate it. 81.147.91.113 (talk) 17:51, 19 April 2009 (UTC)

The conclusion reads:
 * f(x) = (f(x) − g(x)) + g(x)→0 + L = L.

This is not the same as f(x) = L. I don't see anything wrong with the proof. (Maybe a bit too verbose? I like the 1st proof much better.) -- Taku (talk) 21:50, 20 April 2009 (UTC)

Easier proof
I'll use the notation and hypothesis of the article's statement. We need to prove that for any ε>0 there's a δ>0 such that if

0<|x-a|<δ then |f(x)-L|<ε

For every ε, it follows almost imediatelly from the fact that h→L and f→L when x→a that there's a δ>0 such that both


 * f(x)-L|<ε

and


 * h(x)-L|<ε

when |x-a|<δ. On the other hand, the inequalities above imply that -ε<f(x)-L and h(x)-L<ε. Since f(x)≤h(x),

-ε < f(x)-L ≤ h(x)-L < ε

But f(x)≤g(x)≤h(x), so we can squeeze g(x) in the above inequality:

-ε < f(x)-L ≤ g(x)-L ≤ h(x)-L < ε

And therefore:

-ε < g(x)-L < ε

i.e.,


 * g(x)-L| < ε

187.15.3.46 (talk) 00:57, 24 November 2009 (UTC)

Edit Request
The 4th Example is simply WRONG. The example uses the logic that if a < b and c < d  then ac < bd. This is wrong.

a simple counter-example:

1 < 2

-|y| ≤ y

so, according to the example -|y| ≤ 2y;  but for ANY y < 0 this does NOT hold!! (y = -1 and it is false that -1 ≤ -2) I am not even sure that the quotient exists as (x,y) → (0,0). Could someone please check this and if I'm correct fix?173.189.75.215 (talk) 16:26, 27 September 2013 (UTC)
 * Your are right that this logic is faulty for in $$a,b,c,d \in \mathbb{R}$$, but the proof isn't relying on such an argument. First note that the logic is sound for $$a,b,c,d \in \mathbb{R}^+$$ meaning you still can use that argument in a setting where everything is greater than zero. Now to understand the proof, it is probably best to consider the cases $$y \geq 0$$ and  $$y < 0$$ separately to understand how multiplying by $$y \geq 0$$ affects the other inequality in each case and then combine those results using the absolute value notation.--Kmhkmh (talk) 05:07, 30 May 2017 (UTC)
 * I found it very difficult to follow the fourth proof. I believe I got it, but not being a mathematician I think it would be helpful to justify the crucial second step intuitively. I think the idea is that we can multiply by a negative and positive y.  multiplying by negative y will make the expression negative, but not smaller than -abs(y).  Multiplying by positive y will make it positive, but not greater than abs(y).  Therefore we have trapped the expression between -abs(y) and abs(y). The reader is likely to get stuck on the 0 on the left hand side of the inequality. I may add something to the text to this effect.
 * I sure wish Wikipedia entries were a little more user friendly. Danielmabuse (talk) 09:47, 2 January 2024 (UTC)

Redirect from "sandwich theorem"
There is also the sandwich theorem of Lovász. Maybe create a disambiguation page? - Saibod (talk) 13:28, 3 February 2015 (UTC)

circular reasoning
Regarding the second example ((sin x)/x): The formula for the area of a circle is obtained by assuming the limit in question, so it would be better to just go with the lengths indicated in the diagram (that is, sin x, x, and tan x). Also it would be worth noting that since both cos x and (sin x)/x are even functions, that this holds for negative x as well. Kontribuanto (talk) 21:37, 5 March 2024 (UTC)