# Thread: sinx/x and the Squeeze Theorem

1. ## Re: sinx/x and the Squeeze Theorem

Okay. I left this thread open after my last post in the hopes that reason would intrude. I'm not talking about the OP, I'm talking about simple respect. I'm going to say this publicly: No arguments/insults on the open Forum. We have the PM system for that if you insist. And no, Hartlw, I'm not just talking about you. That last note from Prove It was rather nasty as well. I'm not saying you have to agree or even like each other, but if you are doing it in a thread then please stop. I expect you to be professional about it. If you can't then you shouldn't give help.

@Hartlw: Since you seem to be challenging me openly, I will respond in kind as it demonstrates what I am talking about.

The comment about not "locking the thread in time" is completely uncalled for. (There have been a number of comments that fall into this territory.) I have never treated you in any other manner than professional. We don't always agree and you get yourself stuck in a rut and will argue with practically anyone who says something you disagree with so I have, on occasion, been forced to take action against you. This is one of those situations. As I told my students when they had finished taking a test and were getting rowdy "Don't upset the teacher when he's grading your test."

If you want to dare me to mistreat you, then do it over PM.

I'm not locking the thread. Again my hope is that the insults will stop. Hartlw I can't say anything about your posts because I simply don't know the material well enough. But it seems to me like Devano and SlipEternal have made some reasonable points that you can talk about. Otherwise please leave the thread alone.

-Dan

2. ## Re: sinx/x and the Squeeze Theorem

To cut to the chase:
In the squeeze theorem, is g(x)≤f(x)≤h(x), noting that it is standard textbook designation for a closed interval,
a) closed, does not apply to lim sinx/x
b) open, applies to lim sinx/x, but then why not state the theorem as g(x)<f(x)<h(x)
c) ambiguous, depends on f(x), does not apply to lim sin/x

Want to end the thread? Please answer a) b) or c). Explanations OK if not too obtuse, wordy, or irrelevant. This is a simple question.

Allegation of passion is not a mathematical argument, it is a political one (insult). See comment above by Topsquark.

I’m still annoyed at being locked out of the challenge problem, of which this thread is an accidental off-shoot.

3. ## Re: sinx/x and the Squeeze Theorem Originally Posted by Hartlw Allegation of passion is not a mathematical argument, it is a political one (insult). See comment above by Topsquark.

I’m still annoyed at being locked out of the challenge problem, of which this thread is an accidental off-shoot.
Calling someone passionate is not an insult. My comment was in earnest. While I do not agree with your argument, if you are correct, there is a FAR larger issue. Nearly every first year calculus textbook contains an incorrect proof. How could that possibly be an insult to you?

4. ## Re: sinx/x and the Squeeze Theorem Originally Posted by Hartlw Im still annoyed at being locked out of the challenge problem, of which this thread is an accidental off-shoot.
(sighs) You were not locked out of the thread, there's a problem with your account. I have told you this. mash is going to have to handle it and he apparently hasn't been around much recently. No one is out to get you so please stop acting so paranoid.

-Dan

5. ## OP is wrong

1) g(x) ≤ f(x) ≤ h(x) is an inequality that applies for any particular!!! value of x.
2) If g(x) < f(x) < h(x) for any particular!!! value of x, it satisfies 1).
You want ≤ because if a particular f(x) for a particular!!! value of x falls on the edge of the interval, the conclusion isnt invalidated.

I know, I know, x=a can be excluded. Thats a limit issue and not relevant to this discussion. Please don't bring it up againnn.

6. ## Re: sinx/x and the Squeeze Theorem

The squeeze theorem says something about the nature of (real-valued, defined on a real subset) functions in general: If two functions bound a function "in the middle" on an interval (ignore for the moment whether the interval is open or closed, it's an interval), and if this inequality holds even while we "squeeze" the interval to an "arbitrarily" small one (invoking some number like "epsilon" which denotes half the length of our interval), then the limit at the point we squeeze to, of the function in the middle, has to equal the common limits of the bounding functions.

It is helpful, but not necessary, if the bounding functions are continuous. If the bounding functions strictly bound, and are continuous, the squeeze theorem implies the have the same VALUE at the only interval centered at $a$ of length 0: $[a,a] = \{a\}$.

We can apply an inequality like 1) in the post above to an INTERVAL, we say, for example, that $f(x) \leq g(x)$ on $I$, if for every $a \in I$, we have $f(a) < g(a)$.

The interval of relevance here is:

$(-\epsilon,\epsilon) - \{0\})$ (this is actually a "deleted" interval, which we have to use because, unfortunately, $\dfrac{\sin x}{x}$ is not defined at 0), or if you prefer, we consider the two "one-sided intervals":

$(-\epsilon,0)$ and $(0,\epsilon)$.

In this particular case, the two one-sided limits we obtain are equal.

Consider a simpler example, where trigonometry is no longer involved:

Let $h(x) = x$, let $g(x) = -x$, and let $f(x) = \dfrac{0}{x}$.

For all $x \in (0,\epsilon)$ we certainly have: $g(x) \leq f(x) \leq h(x)$ ,and similarly for all $x \in (-\epsilon,0)$, we have: $h(x) \leq f(x) \leq g(x)$.

Is there any doubt that we must have $\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = g(0) = h(0) = 0$?

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