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