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I wrote out a proof of the Squeeze Theorem for my personal notes as I study AP Calculus. This is my first time really writing out a proof using a combination of natural language and logic. I would appreciate feedback on whether there are any errors, style mistakes, or anything I could have stated more cleanly, or anything important I omitted. I'm obviously not submitting this to a journal, rather I'd like to be a good math communicator here on math stackexchange and elsewhere. With that said..

Squeeze Theorem: Suppose that $$ f(x) \leq g(x) \leq h(x)$$ for all $x$ in some interval around $c$, with the possible exclusion of $c$ itself. Also suppose that $$ \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x). $$ Then $$\lim_{x\to c} g(x) = L.$$

Proof: We want to prove $\lim\limits_{x\to c} g(x) = L$ given the premises above it. Let $\epsilon$ be any real number more than $0$, and let $x$ be any real number. By the definition of a limit,

$$ \lim_{x\to c}f(x) = L$$

means there exists a a $\delta_f$ such that,

$$ \begin{align} 0 < |x-c| < \delta_f &\rightarrow |f(x) - L| < \epsilon \\ 0 < |x-c| < \delta_f &\rightarrow -\epsilon < f(x) - L < \epsilon. \tag{1} \end{align} $$

And

$$ \lim_{x\to c}h(x) = L$$

means there exists a a $\delta_h$ such that,

$$ \begin{align} 0 < |x-c| < \delta_h &\rightarrow |h(x) - L| < \epsilon \\ 0 < |x-c| < \delta_h &\rightarrow -\epsilon < h(x) - L < \epsilon. \tag{2} \end{align} $$

The open interval around $c$, with the possible exclusion of $c$ itself, can be described with some $\delta_g$, such that

$$ \begin{align} 0 < |x-c| < \delta_g &\rightarrow f(x) \leq g(x) \leq h(x) \\ 0 < |x-c| < \delta_g &\rightarrow f(x) - L \leq g(x) - L \leq h(x) - L. \tag{3}\\ \end{align} $$

Let $\delta$ be the minimum of $\delta_f$, $\delta_h$, and $\delta_g$. Then by the transitivity of inequality, we can substitute $\delta$ in for the previous deltas in (1), (2), and (3). So with our $\delta > 0$ we have:

$$ \begin{aligned} 0 <|x-c| < \delta \rightarrow& -\epsilon < f(x) - L < \epsilon, \\ & -\epsilon < h(x) - L < \epsilon, \\ & f(x) - L \leq g(x) - L \leq h(x) - L \\ 0 < |x-c| < \delta \rightarrow& -\epsilon < g(x) - L < \epsilon \end{aligned} $$

So by the definition of a limit:

$$ \lim_{x\to c} g(x) = L. $$

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    $\begingroup$ Replace all the $|x-c|<\delta_*$ with $0<|x-c|<\delta_*$, where $\delta^*=\delta, \delta_f,\delta_g,\delta_h$. $\endgroup$ Commented Dec 9, 2021 at 0:11
  • $\begingroup$ Forgot to include that part from definition of a limit, thanks. Updated. $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 0:27
  • $\begingroup$ Why do you have $\delta_g$ here? Just take the minimum of $\delta_f$ and $\delta_h$. $\endgroup$ Commented Dec 9, 2021 at 0:33
  • $\begingroup$ @TedShifrin $ \delta_g $ describes the interval in which $f(x) \leq g(x) \leq (h(x)$. Suppose it is smaller than $\delta_f$ and $\delta_h$, then $\delta$ needs to be that small later, no? $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 0:41
  • $\begingroup$ But that’s just the domain of the functions, so everything happens within that. $\endgroup$ Commented Dec 9, 2021 at 0:46

2 Answers 2

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I would use more symbolic language to make it shorter and clearer/easier to read. For example, cut out the first sentence, then: Let $\epsilon>0$ and $x\in\mathbb{R}$. I would also not mention some other trivial things like “by transitivity of inequalities”. When stating the existence of a $\delta$, I would state there exists $\delta>0$, not just $\delta$ and then having to rely on the inequality $0<|x-c|<\delta$.

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  • $\begingroup$ Is the 𝜖∈R implicit when you just say Let 𝜖>0 ? It wasn't trivial to me that you could replace the deltas on that step. I'd like to reference the right reason. Point noted on $\delta > 0$.. Is the idea there to match the definition of a limit more precisely, even though the implication is obvious? $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 0:55
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    $\begingroup$ It is standard to assume a real-valued variable after you state an inequality like $\epsilon>0$ and say nothing else. saying $\delta>0$ right when you state its existence is important if you want to be very precise, even if later inequalities you write would imply it. I would recommend reading different authors and seeing style differences. There isn’t a single epsilon-delta limit notation. Being 100% aware of why you can take every step is essential, but trivial observations should not be written out if you want to communicate this proof smoothly to others, unless you are teaching. $\endgroup$
    – Suzane
    Commented Dec 9, 2021 at 1:13
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    $\begingroup$ By the way, can I take the opportunity to recommend “Introduction to Analysis” by A. Mattuck? An outstanding book to learn proofs for the first time (and also analysis). $\endgroup$
    – Suzane
    Commented Dec 9, 2021 at 1:24
  • $\begingroup$ Thanks Suzane. Appreciate the feedback, and welcome to Math Stack Exchange assuming you're new! I just now learned that Math Stack Exchange $\neq$ Math Overflow, so I'm editing my question to indicate I'm not at the research or real analysis level yet. Would that book be better to start after calc / linear algebra? $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 1:38
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    $\begingroup$ Thanks! Haha I was just too lazy before to write any answers. I guess it depends on your goals; do you want to learn how to prove stuff? It is not trivial; it takes dedication and good materials. The book I mentioned is the best I know by far for this. And remember analysis is basically proving Calculus results, including limits in the beginning of this book. It would be nice to learn it alongside Calculus. You seem on the right path with wanting to be very precise, good luck. $\endgroup$
    – Suzane
    Commented Dec 9, 2021 at 1:53
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I mostly agree with everything Suzane said. But I would recommend against working with conditions like $0<|x-c|<\delta$ if you can avoid it. It would be clearer to redefine $x$ as a point from a suitable interval, wherever you need it. That is, replace your statement before equation $(1)$ with "By definition, $\lim_{x\to c} f(x)=L$ means there is some $\delta_f>0$ such that for all $x \in (c-\delta_f,c)\cup(c,c+\delta_f)$, we have $$|f(x)-L|<\epsilon,$$ i.e. $$-\epsilon<f(x)-L<\epsilon."$$ (I've never learned a good way to put quote marks after centred equations, don't do that for realsies).

I would also suggest that you be more precise about which inequalities imply which other inequalities. In your second-last block of equations, the reader has to work out for themselves which sides of the first three inequalities they should combine to obtain the final inequality. There are a few options for how you do that: you could say, "From $(3)$ and the first inequality in $(1)$, for all $x \in (c-\delta,c)\cup(c,c+\delta)$, we have $$-\epsilon<f(x)-L\le g(x)-L.$$ Similarly, from (3) and the second inequality in (2), we have $$g(x)-L\le h(x)-L<\epsilon."$$ Alternatively, you don't actually use the fact $f(x)-L<\epsilon$, so you could remove it entirely and just say, "we have $|f(x)-L|<\epsilon$, which implies $-\epsilon<f(x)-L$."

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  • $\begingroup$ Thank you! Appreciate the feedback. One thing I'll note is that we do use the fact $f(x) - L < \epsilon$ in terms of the final step of getting back to my formal definition of a limit, which has $|f(x) - L| < /epsilon$ Do you use the set notation in your definition of a limit? $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 2:14
  • $\begingroup$ I think you'd use $h(x)-L<\epsilon$, but I don't think you use $f(x)-L<\epsilon$. If not, can you explain exactly how you use it? I learned the definition of a limit as "$\lim_{x\to c}f(x)=L$ if and only if, for all $\epsilon>0$, there is some $\delta>0$ such that for all $x \in \mathbb{R} \backslash \{c\}$, $|x-c|<\delta$ implies $|f(x)-L|<\delta$". I would be comfortable saying what I said given my definition, but you could instead say, "The fact $\lim_{x \to c} f(x)=L$ implies that there exists some $\delta>0$, such that for all $x \in (c-\delta,c)\cup(c,c+\delta)$, $|f(x)-L|<L$." $\endgroup$
    – 1Rock
    Commented Dec 9, 2021 at 2:24
  • $\begingroup$ oh sorry, I was confusing f with g(x). yes, we could get rid of parts of the f(x) and h(X) inequalities $\endgroup$
    – Ben G
    Commented Dec 9, 2021 at 4:50

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