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The statement of the Squeeze Theorem as given in Introduction to Real Analysis, 4th Edition by Bartle and Sherbert is :

Let A $\subseteq \Bbb R$, let $f,g,h:A \to \Bbb R$ and let $c \in \Bbb R$ be a cluster point of A. If $$f(x) \leq g(x) \leq h(x)~~\forall~~x\in A,~x\neq c$$ and if $\lim_{x\to c}~f=L=\lim_{x\to c}~h$, then $\lim_{x\to c}~g = L$.

Then, there is a solved example:

$$\lim_{x \to 0}~x^{3/2}=0~(x>0)$$ Let $f(x)=x^{3/2}$ for $x > 0$. Since the inequality $x<x^{1/2}\leq1$ holds for $0 < x \leq1$, it follows that $x^2\leq f(x)=x^{3/2}<x$ for $0<x\leq1$. Since $\lim_{x\to0}~x^2=0$ and $\lim_{x\to0}x=0$,

It follows from the Squeeze Theorem that $\lim_{x\to0}x^{3/2}=0$.

Here is another example from the book:

$$\lim_{x\to 0}~sinx = 0$$ $$-x\leq sinx \leq x~~\forall~~x\geq0$$ Using Squeeze Theorem, we get the desired result.

My question is:

In the statement of the Theorem, it is mentioned that the inequality $f(x) \leq g(x) \leq h(x)$ must hold for all $x \in A$ with $x\neq c$. However, as we see from the first solved example, Squeeze Theorem is applied even though the inequality holds for only a subset $A=(0,\infty)$. Similarly, in the second example, it is applied even though the inequality holds in $[0,\infty)$ instead of $A=\Bbb R$ as the statement says.

Why is this allowed? I am looking for a rigorous proof / explanation.

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  • $\begingroup$ For the first example, you can think $A$ to be the set of real numbers satisfying $ 0 < x \leq 1$. For the second one, $A = (0,\infty) $. This way, squeeze theorem works perfectly. You seem to be confusing the domain of definition of the function. Though it can be define on its natural domain, it can also be define on some subsets of it. $\endgroup$
    – Azlif
    Oct 30, 2019 at 8:40
  • $\begingroup$ @Azlif I see. I was actually hesitant in doing that. But now that I think about it, if we restrict the domain, then so long as $c$ is not an endpoint of the interval restricted to, the limit shouldn't change. Is that why in the second example, you wrote "$A=(0,\infty)$" instead of "$A=[0,\infty)$" even though the inequality does hold at the endpoint $x=0$. $\endgroup$
    – DS2830
    Oct 30, 2019 at 9:12
  • $\begingroup$ My comment on the book is that bad writing is not restricted to novels. $\endgroup$
    – DanielWainfleet
    Oct 30, 2019 at 10:06
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    $\begingroup$ \Bbb R spells $\Bbb R$ which on this site is standard for the set of reals. Similarly \Bbb Z (the integers) and \Bbb N (the natural numbers, which on this site does not include $0$ although I wish it did. We can use \Bbb Z^+ for the positive integers.) $\endgroup$
    – DanielWainfleet
    Oct 30, 2019 at 10:22
  • $\begingroup$ @DanielWainfleet Thanks! I have edited the post accordingly. $\endgroup$
    – DS2830
    Oct 30, 2019 at 10:44

1 Answer 1

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It would be more general and more useful to replace $$\text {If } f(x)\le g(x)\le h(x)\;\forall x\in A,\,x\ne c$$ with the clause $$\text {If } \exists r>0\,\forall x\in (-r+c,r+c)\cap A\,...\;(c\ne x\implies f(x)\le g(x)\le h(x)\,).$$ Convergence is a local property. For any one $d\in A$ with $d\ne c,$ the value ( if any ) of $\lim_{x\to c}g(x)$ is unaffected by the values of $f(d),g(d),h(d)$ (unless there are some additional conditions on $f,g, h$).

The first solved example in the book says "Let $f(x)=x^{3/2}$ for $x>0$". Aside from the inconsistent style ( i.e. in the syntax of the Theorem we should say $g(x),$ not $f(x)\,),$ this implies $A=dom(f)\supset (0,\infty)$ and it is often implicitly assumed to mean $dom(f)=(0,\infty).$ But as you noted, $x^2<x^{3/2}<x$ is false for (e.g.) $x=2\in A.$

But in my "replaced" version, with $A=(0,\infty)$ and $c=0,$ let $r=1.$

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  • $\begingroup$ I see. Do you know of any books or any relevant literature that present this statement? $\endgroup$
    – DS2830
    Oct 30, 2019 at 10:37
  • $\begingroup$ Not that I can name............ $\endgroup$
    – DanielWainfleet
    Oct 30, 2019 at 16:59
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    $\begingroup$ This book has it on Page 104. $\endgroup$
    – DS2830
    Oct 30, 2019 at 17:29

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