Feedback on Proof of Squeeze Theorem 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:
$$ \lim_{x\to c} g(x) = L. $$
 A: 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 $0<|x-c|<\delta$.
A: 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 $-\epsilon<f(x)-L$."
