Is this stronger statement of the squeeze theorem valid? The squeeze theorem formally states that if $f,g$ and $h$ are real functions defined on an interval $I$ containing $c$ as a limit point and satisfy $g(x) \leq f(x) \leq h(x)$ for all $x \in I$ except possibly $c$, and furthermore, $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then  $\lim_{x \to c} f(x) = L$. This is proven by choosing, for given $\epsilon$, the minimum of the two corresponding $\delta$s of $h$ and $g$ as the $\delta$ for $f$. 
This statement, however, isn't as strong as can be. In particular, suppose $g$ and $h$, about every neighbourhood of $c$, "interchange" roles of being the lower and upper bounds of $f$. A concrete example would be $f(x) = 0$, $h(x) = x^2\sin\frac{1}{x}$ and $g = -h$ as $x \to 0$. I think that the conclusion of the squeeze theorem should be valid in this case regardless, because so long as we choose $\delta$ such that $g,h$ are within $\epsilon$ of $L$, $f$ will be within $\epsilon$ of $L$. 
Essentially, I want to weaken the hypothesis so that rather than $\forall x \in I, c \neq x\left[g(x) \leq f(x) \leq h(x)\right]$, it states $\forall x \in I, c \neq x \left[g(x) \leq f(x) \leq h(x) \  \lor h(x) \leq f(x) \leq g(x)\right]$. 

Questions: Is my reasoning correct? If not, what mistake am I making? If so, why is the stronger statement not more common in calculus/analysis texts? 

 A: There is really not more information to be gained. In the situation you describe, you can define new functions $g_1=\min\{g,h\}$, $h_1=\max\{g,h\}$ and use the regular squeeze theorem with $g_1\leq f\leq h_1$. 
On a more practical point of view, in the vast majority of uses of the squeeze theorem the $g$ and $h$ are monotone. 
A: Your reasoning is right.
If, for all $x,$ either $g(x)\le f(x)\le h(x)$ or $h(x)\le f(x)\le g(x),$ then $$\min\{g(x),h(x)\}\le f(x)\le\max\{g(x),h(x)\}$$ for all $x.$ So all we have to do is show that, if $g(x)\to L$ and $h(x)\to L,$ then $\max\{g(x),h(x)\}\to L.$ (Of course it follows that $\min\{g(x),h(x)\}=-\max\{-g(x),-h(x)\}\to L$ as well.)
Observe that
$$\max\{g(x),h(x)\}=\frac{g(x)+h(x)+|g(x)-h(x)|}2\to\frac{L+L+|L-L|}2=L.$$
A: Your weakened hypothesis is equivalent to $\min(g(x), h(x)) \leq f(x) \leq \max(g(x), h(x))$. So, define $m(x) = \min(g(x), h(x))$ and $M(x) = \max(g(x), h(x))$ and the hypothesis $m(x) \leq f(x) \leq M(x)$ applies. So, is your version really more general?
