Proving the squeeze theorem I know there are posts out there asking about this, but I didn't want to look at other solutions before I feel like I've solved it on my own or at least gotten close. If I'm on the wrong track, any hints would be very welcome instead of a solution. Here's what I have:

Prove the suqeeze theorem: suppose $g(x) \leq f(x) \leq h(x)$ for all $x\in\mathbb{R}$ satisfying $0<|x-c|<\delta$ for some $\delta > 0$. If $\lim_{x\rightarrow c} g(x) = L = \lim_{x\rightarrow c} h(x)$, then $\lim_{x\to c} f(x) = L$.

Proof:
By definition, for $0 < |x-c| < \delta$, $\delta > 0 \implies |h(x)-L| < \epsilon$, $|g(x)-L| < \epsilon$ for any $\epsilon > 0$. Then,
$$ \begin{align}
|h(x) - g(x)| - |g(x) - L| &\leq |h(x) - g(x) + g(x) - L| && |h(x) - L| < \epsilon  \\
|h(x) - g(x)| - |g(x) - L| &< \epsilon && \text{limit definition}\\
|h(x) - g(x)| &\lt \epsilon + |g(x) - L| && \text{rearrange} \\
|h(x) - g(x)| &< 2\epsilon && \text{limit definition}
\end{align} $$
Now, since $g(x) \leq f(x) \leq h(x)$, then $|g(x) - f(x)| \leq |h(x) - g(x)| $. This means that $|g(x) - f(x)| < 2\epsilon$. Then,
$$ \begin{align}
|f(x) - L| - |g(x) - f(x)| &\leq |g(x) - f(x) + f(x) - L|  && \text{triangle rule} \\
|f(x) - L| - |g(x) - f(x)| &< \epsilon && |g(x) - L | < \epsilon\\
|f(x) - L| &\lt \epsilon + |g(x)-f(x)| && \text{rearrange} \\
|f(x) - L| &< 3\epsilon && |g(x) - f(x) < 2\epsilon|
\end{align} $$
I could start with $\epsilon/3$ to end with $|f(x)-L| < \epsilon$; regardless, is this proof correct? Thank you!
 A: Your first sentence is not correct, though I understand what you mean. I think you should say something like: "Let $\varepsilon > 0$ be arbitrary. By making $\delta$ smaller if necessary, we may assume that for all $x \in \mathbb{R}$, $$0 < |x - c| < \delta \implies |h(x) - L| < \epsilon \text{ and } |g(x) - L| < \varepsilon."$$
The rest of your math is correct, but you should preface the computations with something like "let $x \in \mathbb{R}$ with $0 < |x - c| < \delta$ be arbitrary" because you never said what $x$ was.
The proofs of the inequalities seem a bit unnatural. In my opinion, it is more natural to do
\begin{align}
|h(x) - g(x)| &\leq |h(x) - L| + |L - g(x)| \\
&< 2\varepsilon.
\end{align}
and
\begin{align}
|f(x) - L| &\leq |f(x) - g(x)| + |g(x) - L| \\
&\leq |h(x) - g(x)| + |g(x) - L| \\
&< 3\varepsilon.
\end{align}
A: This proof looks just about fine. Here is one comment I have:

By definition, for $0 < |x-c| < \delta$, $\delta > 0 \implies |h(x)-L| < \epsilon$, $|g(x)-L| < \epsilon$ for any $\epsilon > 0$.

It is unclear whether or not you are taking $\delta$ to be the minimum of the $\delta$'s coming from $h$ and $g$ for any $\varepsilon>0$. Personally, I would make that more explicit if that is what you intended.
To shorten up the part where you show that $|h(x)-g(x)|<2\varepsilon$, I would make use of the fact that $\lim\limits_{x\to c}h(x)=L=\lim\limits_{x\to c}g(x)$. Then $$|h(x)-g(x)|\leq |h(x)-L|+|L-g(x)|<2\varepsilon $$ for $x\in\mathbb{R}$ with $0<|x-c|<\delta$ and $\varepsilon>0$.
Other than that the proof is correct.
Again, to cut down on some time, here is an alternate proof:

Let $\varepsilon>0$. Let $x\in\mathbb{R}$ with $0<|x-c|<\delta$, where $\delta$ is the minimum of the $\delta$'s coming from $g$ and $h$. Then we have
\begin{align}-\varepsilon<-|g(x)-L|\leq g(x)-L\leq f(x)-L\leq h(x)-L\leq |h(x)-L|<\varepsilon\end{align}
which implies $|f(x)-L|<\varepsilon$.

