Using inequalities and limits Is it possible to say:
$$
If \ f(x) \ and \ g(x) \ both \ have \ limits \ as \ x\to p\ and \ f(x) \le g(x), \ then \lim_{x \to p} f(x)\le \lim_{x \to p} g(x).
$$
My proof(Edit: Proof is wrong due to assumption of the conclusion):
$$
Assume \ H(x)=f(x)-g(x) \ ,then\\ \lim_{x \to p} H(x)=\lim_{x \to p}[f(x) - g(x)]\\=\lim_{x \to p}f(x)-\lim_{x \to p}g(x) \\\\so \ \lim_{x \to p}f(x)\le \lim_{x \to p}g(x)\ \square
$$
If this theorem is true is it possible to prove the Squeeze Principle(without $\epsilon,\delta$):
$$
Suppose \ that  \ f(x)\le g(x)\le h(x) \ for \ all x\ne p \ in \ some\ neighborhood \ N(p). \ Suppose \ also \ that \lim_{x\to p}f(x)=\lim_{x\to p}h(x) =a. \\Then\ we\ also\  have\ \lim_{x\to p}g(x)=a. 
$$
My proof:
$$
\lim_{x\to p}f(x)\le\lim_{x \to p}g(x) \ \ and \ \lim_{x \to p}g(x)\le \lim_{x \to p}h(x)\ so\\a\le \lim_{x \to p}g(x) \ and \ \lim_{x \to p}g(x) \le a  \ so\\\lim_{x \to p}g(x)=a\ \square.
$$
 A: Suppose that $\lim_{x \to p}f(x) = A, \lim_{x \to p}g(x) = B$ and we know that $f(x) \leq g(x)$ in neighborhood of $p$. Now we need to understand the meaning of the limit statements informally. The values of $f(x)$ are near to $A$ and values of $g(x)$ are near to $B$ when $x$ is near $p$. Now on the contrary suppose that $A > B$. First let us understand this by setting $A = 2, B = 1$. Now values of $f(x)$ are near $A = 2$ so for example like in range $1.9-2.1$ when $x$ is near $p$ and values of $g(x)$ are near $B = 1$ for example in range $0.9-1.1$ Clearly we can see that $$0.9 < g(x) < 1.1 < 1.9 < f(x) < 2.1$$ and this contradicts that $f(x) \leq g(x)$ when $x$ is in a certain neighborhood of $p$. Hence we must have $A \leq B$.
A formal proof consists of showing the same argument without assuming any specific values of $A, B$ (like $A = 2, B = 1$ of previous paragraph). Again let's assume that $A > B$ and let $C = (A + B)/2$ be number lying exactly in between $A$ and $B$ ($B < C < A$). Now idea is to keep values of $g(x)$ near to $B$ such that they don't exceed $C$ and values of $f(x)$ near to $A$ so that they always exceed $C$. This will make $g(x) < f(x)$ contrary to given condition.
Now we put the epsilon delta stuff. Setting $\epsilon = (A - B)/2$ so that $C = B + \epsilon = A - \epsilon$ we see that we have a $\delta > 0$ such that $|f(x) - A| < \epsilon$ for $0 < |x - p| < \delta$ and $|f(x) - B| < \epsilon$ for $0 < |x - p| < \delta$. Thus we have for $0 < |x - p| < \delta$, $$B - \epsilon < g(x) < B + \epsilon = C = A - \epsilon < f(x) < A + \epsilon$$ and thus $g(x) < f(x)$ and we arrive at a contradiction. Hence $ A \leq B$.
The above explanation shows that an epsilon delta argument is not difficult to give provided we really understand the meaning of an informal argument clearly (its more like a translation into an unambiguous language).
As to the case of Sandwich Theorem (Squeeze theorem) with $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to p}f(x) = \lim_{x \to p}h(x) = L$ it is easy to show that $\lim_{x \to p}g(x)$ exists. Corresponding to any $\epsilon > 0$ we have a $\delta > 0$ such that $$L - \epsilon < f(x) < L + \epsilon, L - \epsilon < h(x) < L + \epsilon$$ whenever $0 < |x - p| < \delta$. Hence we get $$L - \epsilon < f(x) \leq g(x) \leq h(x) < L + \epsilon$$ i.e. $L - \epsilon < g(x) < L + \epsilon$ so that $\lim_{x \to p}g(x)$ exists and is equal to $L$. You see, you don't need to use the previous result (dealing with inequalities of two functions and their limits) to get the Sandwich theorem.
A: It is true that if $\lim_{x \to x_0} f(x) = a$ and $\lim_{x \to x_0} g(x) =b$ and for every $x$ it holds that $f(x) \leq g(x)$, then $a \leq b$. And as your proof would show, this is equivalent to: if for every $x$ it holds that $f(x) \geq 0$ and if $\lim_{x \to x_0}f(x) = a$, then $a \geq 0$. Though I do think you are assuming the latter claim to prove the former (ie. in your proof, how do you know that $\lim_{x \to x_0}H(x) \geq 0$?). Another approach would be to assume that $f(x) \geq 0$ for every $x$ and $\lim_{x \to x_0} f(x) = a$, but $a < 0$. Draw a contradiction by using the definition of convergence and cleverly setting $\epsilon = |a|/2$. You should find that with $a < 0$ you must have some $x$ near $x_0$ such that $|f(x)-a|\leq |a|/2 \implies f(x) <0$ contradicting the assumption.
For the squeeze theorem, IF you know that $\lim_{x \to x_0}g(x)$ exists, then your proof is good. However,  you might want to a little more work to argue that the limit of $g(x)$ exists as $x \to x_0$.
