# sinx/x and the Squeeze Theorem

#### Hartlw

Can the squeeze theorem be modified to f(x)<g(x)<h(x) and lim f(x) =L and lim g(x) = L?
But that rules out the situation where you can only show <=, >=.

#### Deveno

MHF Hall of Honor
A STRICT bound is also any LOOSER bound. If we show $x$ is between (but not equal) to $a$ and $b$ (with $a < b$), it follows that one of the following statements is true:

1. $x$ is between, but not equal to, $a$ and $b$
2. $x$ is either $a$ or $b$.

Of course we KNOW that it's the first one, and not the last one, but 1 is in the set {1} U {2}. The statement:

$a \leq x \leq b$ is just a handy abbreviation for the statement (1) or (2) or both (and (2) or both is just extra baggage, but it doesn't hurt anything).

We can't EXTEND statements about $(a,b)$ to $[a,b]$, but GIVEN a statement which is true on $[a,b]$, we can surely say in holds for the RESTRICTION to $(a,b)$.

In other words: GIVEN the $\leq$ as a pre-condition, we can safely drop the "or equal to" and just use $<$. Given ONLY $<$, we CANNOT assume it holds when we use $\leq$ instead.

You COULD make an argument for a "modified squeeze theorem" where the three functions are always strictly unequal, except at the limit point. But this would limit the applicability:

Sometimes, a function oscillates between two bounding functions, coming in contact with them every so often, like:

$f(x) = x^2\sin\left(\dfrac{1}{x}\right)$

which will touch the parabolas $y = x^2$ and $y = -x^2$ at many points that get closer together as we approach 0. However, we always have:

$-x^2 \leq x^2\sin\left(\dfrac{1}{x}\right) \leq x^2$

and the outer two functions have limit 0, as x approaches 0.

The values of this function are EXTREMELY difficult to calculate when $|x|$ is very small, and in point of fact, the derivative at 0 does not exist (it's literally trying to be "two places at once", something a function cannot achieve). In fact it DOES NOT MATTER what value we assign to:

$f(x) = x^2\sin\left(\dfrac{1}{x}\right)$

at $x = 0$, there is NO value that makes it differentiable at 0, although defining $f(0) = 0$ DOES make it continuous.

By the way, this is a standard example that used to be taught in any first-year calculus class.

I do not think you have a clear grasp of logic, and how it pertains specifically to subsets of the real numbers. I'd advise you to stay away from metric spaces entirely, you'll be hopelessly confused.

**********************************

Since $\dfrac{\sin x}{x}$ has a LIMIT at 0, we can DEFINE for $f(x) = \dfrac{\sin x}{x}$ the CONTINUOUS function:

$g(x) = f(x), x \neq 0$
$g(0) = 1$

Personally, I find the "geometric arguments" unconvincing. The reason being: first of all, $\sin(x)$ should be a continuous, and differentiable function. For analysis, this definition should be ANALYTIC-why bother going through all the rigamarole of Cauchy sequences of rational numbers, if at the middle, you are going to stop and refer to the plane, whose algebraic properties are not even PROVEN. What does "length" of a chord, or "area" of a triangle even MEAN?

For example, children are taught that the area of a triangle is $\dfrac{1}{2}bh$. Why is this true?

It is possible, using ONLY calculus, to define a function that has all the properties of "sine" that we need. BUT....it's not pretty, and it would take me far too long to post it here. And some other "big gun theorems" have to be proved first: namely, the Inverse Function Theorem, which is rather complicated. Finally, we need an ANALYTIC definition of the real number $\pi$, because radian measure ultimately depends on arc-length, which poses technical problems of its own (it is not a trivial matter to prove the circle is a "rectifiable curve", as the so-called "troll proof that $pi$ = 4" highlights).

Geometry, ultimately, deals with "constructible numbers", and showing that these numbers are a sub-field of the real numbers, and that $\pi$ is not one of them, is WAY beyond the level of most undergraduate students, except perhaps at the very best math departments in the country. It certainly is very far afield from the matter at hand.

It is, perhaps, defensible to use the series definitions of the exponential function, and trigonometric functions, as a starting point. Doing this would have the unfortunate side-effect of delaying some of the slicker integration techniques (as well as the differentiation of a good many functions) until the very end of the class. Showing that these functions really ARE the ones we have always known and loved, would be yet another safari-length excursion.

So the "geometric" or as I call them "hand-waving" proofs are trotted out, and the gullible students take them as fact, for the indoctrination that teacher is always right, has long since reached the critical point.

• 2 people

#### Hartlw

If x ≠ 0, does sinx/x satisfy cosx ≤ sinx/x ≤ 1/cosx? Yes. Why?
2 ≤ 4
The squeeze theorem proves lim sinx/x=1

I assume from all the bloviation and insults that the other respondents didn’t really understand this.

Now could somebody following this thread please do me a favor and post my post #14 in:
http://mathhelpforum.com/math-challenge-problems/229520-l-hospital-s-rule.html

to
http://mathhelpforum.com/calculus/229484-challenging-limit.html ,
where it properly belongs (the correct answer) which won’t allow me access?

Thanks

It really isn’t fair to post a challenge question, thank the incorrect answers, and then deny access to me with the correct answer.

#### topsquark

Forum Staff
It really isn’t fair to post a challenge question, thank the incorrect answers, and then deny access to me with the correct answer.
It was never a challenge problem. I had wanted it to be...it was just about perfect: A simple looking limit that could not (probably) be done by elementary means unless you knew how to approach it. As I myself was unable to do it I decided not to use it as a challenge problem.

-Dan

#### Hartlw

It was never a challenge problem. I had wanted it to be...it was just about perfect: A simple looking limit that could not (probably) be done by elementary means unless you knew how to approach it. As I myself was unable to do it I decided not to use it as a challenge problem.

I came up with a nice little limit for the Challenge Problems section. (From original thread, hartlw)

-Dan
And yet you thanked people who purportedly posted the right answer (it wasn't).

So why won't you post my (correct) answer? It takes a few seconds.

#### topsquark

Forum Staff
And yet you thanked people who purportedly posted the right answer (it wasn't).

So why won't you post my (correct) answer? It takes a few seconds.
Frankly because I could understand what was being said, but it is far from my area of expertise. However from what little I know the others are correct.

If I see a solution I feel is elegant I tend to thank them. I have "backed the wrong horse" before but I'm not worried about it and I have no intention of changing my thanking "policies." By thanking others I am implying they have a good argument, not that they are correct in all respects. I have also not thanked a number of members that are correct in their arguments. I may thank you later on in this thread or I may not. That is not always a statement of whether or not I think you are correct.

The thanks system is not there for validation of correct answers...It is a measure of respect for the work someone else has done. You are taking my not thanking you in this thread as an insult...that's not what is happening, but it's your call to make.

-Dan

#### SlipEternal

MHF Helper
I am not sure I follow the exact nature of the dispute, and I am not trying to weigh in. I only offer a possible alternative way of looking at the original problem that I hope will help everyone reach consensus. The original problem (if I am following this thread correctly) was to find $$\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}$$. There seems to be some disagreement over when x may or may not take on certain values (specifically zero). The following function is continuous over all real numbers. Therefore, $$\displaystyle \lim_{x \to 0} f(x) = f(0) = 1$$, and the proof of its continuity is likely equivalent to a proof that $$\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$$:

Define $$\displaystyle f:\Bbb{R}\to \Bbb{R}$$ by $$\displaystyle f(x) = \begin{cases}\dfrac{\sin x}{x} & x\neq 0 \\ 1 & x=0\end{cases}$$

Hartlw, would using this function satisfy your concerns for using the Squeeze Theorem? Now, the function is defined at $$\displaystyle x=0$$, and in fact takes on the value we expect the limit to be at $$\displaystyle x=0$$. Alternately, if this does not help in the use of the Squeeze Theorem, does it help reinforce your point somehow that the Squeeze Theorem does not apply?

#### Prove It

MHF Helper
You seem to be concerned about the "hypothesis of the squeeze theorem being satisfied". Many people actually get the definition of a limit incorrect. They think:

$\displaystyle \lim_{x \to a} f(x) = L$ means:

"For any $\epsilon > 0$ there exists a $\delta > 0$ such that $|x - a| < \delta \implies |f(x) - L| < \epsilon$".

This is absolutely wrong, the correct definition is:

"For any $\epsilon > 0$ there exists a $\delta > 0$ such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$".

The interval $(a -\delta,a) \cup (a,a+\delta)$ is what is intended for $x$ to lie in: this is also called a "deleted neighborhood" of $a$.

The formal statement of the squeeze theorem does not require $g(x) \leq f(x) \leq h(x)$ for ALL points in a neighborhood of $a$ ($a = 0$ in this case), but rather all points in a deleted neighborhood of $a$.

Note that if, in fact, $g(x) < f(x) < h(x)$ in said deleted neighborhood (that is, the inequality is strict) this implies the (looser) restriction $g(x) \leq f(x) \leq h(x)$.

Let's prove the simple statement: $p < q \implies p \leq q$ via truth-tables. First off, note that $p \leq q$ is shorthand for the compound statement:

$(p < q) \vee (p = q)$.

Now, if $p = q$ or $p > q$, the implication is true, no matter if the consequent is true or false (a false statement can imply anything). So, without loss of generality, assume $p < q$ is true. Then the implication is true if and only if the consequent is true.

Since the consequent is an "or" statement, it is true if either $p = q$ is true, or $p < q$. If $p < q$, then $p = q$ is false and $p < q$ is true. So our implication has the form:

T --> (F or T), equivalent to:
T --> T, equivalent to:
T

Your objection that $f$ continuous on $(a,b)$ does not mean $f$ is continuous on $[a,b]$ is totally spurious. It is OBVIOUS that $(a,b) \subseteq [a,b]$, and therefore if $x \in (a,b)$, then $x \in [a,b]$, by the DEFINITION OF SUBSET.

This does NOT mean $[a,b]$ has every property $(a,b)$ has, you have this exactly BACKWARDS: for example $b$ is in the set $[a,b]$, but it most assuredly NOT in $(a,b)$.

In short, ProveIt is ABSOLUTELY CORRECT. It is you who are confused, and I suggest you actually LEARN some mathematics before pontificating about it.
Thank you Deveno. There is just no reasoning with some people. Can I suggest to the mods to lock this thread please? And I'd also suggest to anyone else who sees this thread - never argue with an idiot, they will bring you down to their level and beat you with experience...

#### Hartlw

If g(x) is monotonically increasing and h(x) is monotonically decreasing, the squeeze theorem is a nest. The nest defines a unique limit if the intervals are closed. That is not the case for sinx/x, which is not closed because chord length can never equal arc length. OP is correct.

If you do decide to lock me out of this discussion, please have the decency to announce it in the thread. I assume it'll happen after the next bloated irrelevant thread which buries this one. By the way, if you write down 100 things that are true it doesn't necessarily prove your point.

As for the challenge problem, sin x is only defined analytically by a series so the answer is lim(x-sinx)/x3=6 can't be proven under the conditions of the challenge. If you're not that familiar with it, why are you judging who is correct and who is incorrect?

#### SlipEternal

MHF Helper
If g(x) is monotonically increasing and h(x) is monotonically decreasing, the squeeze theorem is a nest. The nest defines a unique limit if the intervals are closed. That is not the case for sinx/x, which is not closed because chord length can never equal arc length. OP is correct.

Hartlw, you are clearly passionate about this. Nearly every first year Calculus textbook I have seen includes some form of the proof of this very limit: $$\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$$ using geometric interpretation and the Squeeze Theorem. Perhaps you should write your concerns to publishers. If nearly every Calculus textbook written contains an incorrect proof, that is a far more serious matter than people on this forum disagreeing with you, isn't it?