Talk:Squeeze theorem

message: -->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)

message: 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)
thread_title:  "Kathy Theorem" 

message: :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)
replies: text:  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)
replies: text: 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)
thread_title:  Italian name 

message: 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)
replies: text: 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)
thread_title:  Real world uses 

message: 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)
replies: text: 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)
thread_title:  "Proof" 

message: 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)
thread_title:  Easier proof 

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

message: 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)
thread_title:  circular reasoning