Is this a valid proof of the squeeze theorem? I'm self-studying Spivak's calculus, and I have no way of checking my solutions. One of the problems asks for a proof of the squeeze theorem. Here is what I have figured: 
Proof. Suppose there exist two functions $f(x)$ and $g(x)$ such that $(\forall x)(g(x) \geq f(x))$ and both $\lim _{x\to a}f(x) = l_1$ and $\lim _{x\to a}g(x) = l_2$ exist. Now let $h(x) = g(x) - f(x)$ so that $h(x) \geq 0$. Assume that $l_2$ $<$ $l_1$. 
By previous results, $(\forall \epsilon > 0)(\exists \delta > 0): (\forall x)($if $\lvert x - a \rvert < \delta$, then $\lvert h(x) - (l_2 - l_1) \rvert < \epsilon)$. 
But if $\epsilon \leq l_1 - l_2$, no $\delta$ exists such that $\lvert h(x) - (l_2 - l_1) \rvert < \epsilon$, thus, a contradiction. 
$\therefore l_2 \geq l_1$ 
Given this result, define three functions $f_1(x)$, $f_2(x)$, and $f_3(x)$ such that $(\forall x)(f_1(x) \leq f_2(x) \leq f_3(x))$ and $\lim _{x \to a}f_1(x) = \lim _{x \to a}f_3(x) = L$. 
Because $f_2(x) \geq f_1(x)$, it must be true that $\lim_{x \to a}f_2(x) \geq \lim_{x \to a}f_1(x)$ and similarly, because $f_3(x) \geq f_2(x)$, it must also be true that $\lim_{x \to a}f_3(x) \geq \lim_{x \to a}f_2(x)$. 
It follows that $\lim_{x \to a}f_1(x) \leq \lim_{x \to a} f_2(x) \leq \lim_{x \to a} f_3(x)$, which in turn makes true that $L \leq lim_{x \to a} f_2(x) \leq L$. 
$\therefore \lim_{x \to a}f_2(x) = L$ $\blacksquare$
I'm mostly unsure about the validity of the first part, specifically the "no $\delta$ exists" bit. Also, I would appreciate suggestions regarding formal proof-writing as I'm new to this. Thanks a lot in advance! 
 A: Why does no such $\delta$ exist? This is not immediately obvious. you need to explain it.
A second problem with your proof is that the squeeze theorem only assumes that $f_1$ and $f_3$ converge. That the squeezed function $f_2$ also has a limit is part of what must be proved. But your proof simply assumes this is so.
For a better approach, consider this: if $\epsilon > 0$, then there is a $\delta_1 > 0$ such that for $|x - a| < \delta_1$, $f_1(x) > L - \epsilon$. And there is a $\delta_2 > 0$ such that for $|x - a| < \delta_2$, $f_3(x) < L + \epsilon$.
A: There are two results which you are trying to mix up:
Theorem 1: Let $f, g$ be functions defined in a deleted neighborhood $I$ of $a$ such that $f(x) \leq g(x)$ for all $x \in I$. If both the limits $L_{1} = \lim_{x \to a}f(x), L_{2} = \lim_{x \to a}g(x)$ exist then $L_{1} \leq L_{2}$.
Thereom 2 (aka Squeeze Theorem): If $f, g, h$ are functions defined in a deleted neighborhood $I$ of $a$ and $g(x) \leq f(x) \leq h(x)$ for all $x \in I$ and both the limits $\lim_{x \to a}g(x), \lim_{x \to a}h(x)$ exist and are equal to $L$ then $\lim_{x \to a}f(x)$ also exists and is equal to $L$.
The first theorem holds true even when $f(x) \leq g(x)$ is replaced by $f(x) < g(x)$. This is the first part which you have proved. The proof you have given is correct, but it does not show very clearly that you have understood it correctly. In fact you need to get a contradiction by assuming that $L_{1} > L_{2}$. The way to obtain contradiction is to use an $\epsilon$ such that $0 < \epsilon < L_{1} - L_{2}$ and thereby get a $\delta > 0$ such that $$L_{2} - L_{1} - \epsilon < g(x) - f(x) < L_{2} - L_{1} + \epsilon < 0$$ for all $x$ with $0 < |x - a| < \delta$. This contradicts that $g(x) \geq f(x)$ for all $x \in I$.
Another mistake (which many students make) is "trying to derive theorem 2 from theorem 1". This is not possible because theorem 1 requires the existence of limit for functions on both sides of an inequality. In theorem 2 we have two inequalities $g \leq f$ and $f \leq h$ and clearly we don't know anything about existence of limit of $f$ so theorem 1 does not apply to any of these inequalities. Theorem 2 is simply independent of theorem 1 and has an easier proof.
By given hypothesis in the theorem 2, it follows that for every $\epsilon > 0$ it is possible to find $\delta_{1} > 0, \delta_{2} > 0$ such that $$L - \epsilon < g(x) < L + \epsilon\tag{1}$$ for all $x$ with $0 < |x - a| < \delta_{1}$ and $$L - \epsilon < h(x) < L + \epsilon\tag{2}$$ for all $x$ with $0 < |x - a| < \delta_{2}$.
If we set $\delta = \min(\delta_{1}, \delta_{2})$ then both the inequalities $(1), (2)$ hold simultaneously for all $x$ with $0 < |x - a| < \delta$ and hence we have $$L - \epsilon < g(x) \leq f(x) \leq h(x) < L + \epsilon$$ for all $x$ with $0 < |x - a| < \delta$. This means that $$L - \epsilon < f(x) < L + \epsilon$$ for all $x$ with $0 < |x - a| < \delta$ and thus $\lim_{x \to a}f(x)$ exists and is equal to $L$.
Note that the power (and usefulness) of Squeeze theorem is entirely due to the fact that it gives us the existence of limit of $f$ from the existence of limit of $g, h$. In practice it is possible to establish inequalities of type $g \leq f \leq h$ where $f$ is a complicated or unfamiliar function whereas $g, h$ are simpler/familiar functions whose limits are known and thereby we can evaluate the limit of $f$.
