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?