sinx/x and the Squeeze Theorem

Prove It

MHF Helper
Yes, but sinx/x is not defined for x=0. I note that wiki gets squeeze theorem and inequality right (only<), but draws the wrong conclusion, (Lim = 1 by squeeze theorem) the subject of OP.
I suspect your knowledge of limits is limited. A function does NOT have to be defined at a point in order for it to have a limit there.

In fact, if it WAS the case that a function had to be defined at a point in order to have a limit at that point, there would be no such thing as a derivative, since the derivative is defined as \displaystyle \begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \end{align*}, even though the function is NOT defined at h = 0.

As another example, consider the limit \displaystyle \begin{align*} \lim_{x \to 0}\frac{x^2 + x}{x} \end{align*}. If you drew a graph of the function, you would see that it is identical to \displaystyle \begin{align*} x + 1 \end{align*}, except that it has a hole where x = 0. But the function still approaches the same value as you close in on the point x = 0 from both directions. You would most likely have been taught to factorise and cancel, and then substitute the value in. This is fine, in fact it's exactly what you are supposed to do, and you would find that the limiting value is 1. But the function is NOT defined there.

More specifically, the precise definition of a limit is that if you can show for all \displaystyle \begin{align*} \epsilon > 0 \end{align*} that there exists a \displaystyle \begin{align*} \delta > 0 \end{align*} such that \displaystyle \begin{align*} 0 < \left| x - c \right| < \delta \implies \left| f(x) - L \right| < \epsilon \end{align*} then it proves \displaystyle \begin{align*} \lim_{x \to c} f(x) = L \end{align*}. Notice the fact that \displaystyle \begin{align*} 0 < \left| x - c \right| \end{align*}, this literally means that there has to be some distance between x and c, which means that \displaystyle \begin{align*} x \neq c \end{align*}, thereby enabling the function to not need to be defined at \displaystyle \begin{align*} x = c \end{align*} in order for there to be a limit there.

Hartlw

A function does NOT have to be defined at a point in order for it to have a limit there. /QUOTE]

Correct, that is why I wonder that you keep using sinx/x, x=0 (incorrect) in your hypothesis.

Perhaps it would help to see it done correctly, not using the squeeze theorem (incorrectly):

Just out of curiosity, I thought I’d look for someone who gets it right, not that easy. Courant, Dif and Int Calc vol 1 pg 48, does:
Using the same figure you do, but a slightly different lower area based on chord instead of the triangle vertical (which is tighter), he correctly gets:

“1 < x/sinx < 1/cosx”

And correctly concludes:

“We know that cos x tends to 1 as x→∞ and from this it follows that the quotient sinx/x can differ only arbitrarily little from 1, provided that x is near enough to 0. This is exactly what is meant by the equation which was to be proved.” (lim sinx/x=1).

That is a correct conclusion, not the squeeze theorem.

Prove It

MHF Helper
Well first of all, \displaystyle \begin{align*} \cos{(x)} \end{align*} does NOT go to 1 when \displaystyle \begin{align*} x \to \infty \end{align*} (in fact, that limit is undefined), but I suspect that is a typo and should be 0.

And now I understand that the reason you are having so much trouble with the squeeze theorem in this case. It's because your definition is weak.

The FULL definition of the squeeze theorem is this:

The squeeze theorem is formally stated as follows.

Let I be an interval having the point a as a limit point. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have:

$\displaystyle g(x) \leq f(x) \leq h(x)$

and also suppose that:

$\displaystyle \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$

Then $\displaystyle \lim_{x \to a} f(x) = L$.

The functions g and h are said to be lower and upper bounds (respectively) of f.

Here a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits.

Hartlw

The squeeze theorem is formally stated as follows.

Let I be an interval having the point a as a limit point. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have:

$\displaystyle g(x) \leq f(x) \leq h(x)$

so a=0 disallowed (sinx/x not defined for x=0), in which case:

g(x)<f(x)<h(x) and condition of squeeze theorem not fulfilled.

I simply can't understand why you insist on using x=0 in sinx/x, contary to your own definitions and arguments.

Prove It

MHF Helper
That's not what it's saying at all. Is English your first language?

It is saying that as long as the inequality holds for all x in the domain specified EXCEPT POSSIBLY AT THAT POINT then it is enough to prove the limit.

I am not saying anything at all about the value of sin(x)/x at x = 0. You are correct, it is undefined there. But the function DOES NOT NEED TO BE DEFINED THERE.

And surely if g(x) < f(x) < h(x) at all points except x = a, then \displaystyle \begin{align*} g(x) \leq f(x) \leq h(x) \end{align*} for all points there too.

Do you understand that "less than" implies "less than or equal to"?

Hartlw

That's not what it's saying at all. Is English your first language?

Do you understand that "less than" implies "less than or equal to"?
Absoluteley incorrect, and may be the source of your confusion.

Let fx) be continuous on a<x<b. This means f(x) continuous on a<=x<=b???

--------

Could anybody 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

Prove It

MHF Helper
First of all, I am not the one who is confused. Your entire argument about the squeeze theorem being inappropriate is a matter of semantics from your poor definition of the squeeze theorem. The full definition gets around this problem, the squeeze theorem is appropriate and any further argument is pointless.

The other point I was making was that \displaystyle \begin{align*} a < b \implies a \leq b \end{align*}. So if \displaystyle \begin{align*} g(x) < f(x) < h(x) \end{align*} at all x except the limiting point a, then surely \displaystyle \begin{align*} g(x) \leq f(x) \leq h(x) \end{align*}.

Hartlw

As with all your other arguments your last one is incorrect. You are:
1) Ignoring what I am saying.
3) Contradicting yourself. One minute you are saying x=0 is ok and the next that it isn't.

But thanks for emphasizing the significance and importance of my OP, and confirming it's correctness with the intensity, ferocity, and irrationality of your attacks.

Deveno

MHF Hall of Honor
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.

1 person

Hartlw

Here we go again.

Argument by obfuscation.

The definition of a limit is in any standard calculus textbook. "Some people think.." is nothing but an underhanded insult (straw man).

The notion that a<f(x)<b is the same as a<=f(x)<=b is of course nonsense.
Only < > imply for lim sinx/x for the simple reason that a chord has to be less than an arc, it can't be <=.

If a theorem requires f(x) to be continuous on [a,b] is it sufficient that it be continuous on (a,b)?

But while I have your attention Deveno,

Could you 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