Squeeze Theorem with strict inequalities Suppose that $I$ is an interval, and $g,f,h$ are functions defined on $I$, except possibly at $x_0$. Furthermore, $\forall \varepsilon>0,$ $\forall x\in I,$ and for some constant $L$,
$$ (L-\varepsilon)g(x) < f(x) \le Lh(x) .$$
Also,
$$\lim_{x\to x_0} g(x) = \lim_{x\to x_0} h(x) = 1.$$
Then is it true that $\forall \varepsilon>0,$
$$L-\varepsilon < \lim_{x\to x_0} f(x) \le L,$$
and since $\varepsilon$ can be taken to be arbitrarily small, the middle is forced to give
$$\lim_{x\to x_0} f(x)=L?$$
Note that this is essentially the Squeeze Theorem with a strict lower-bound.
 A: Suppose that $L$ is fixed. The quantified statement $$(\forall \epsilon > 0) (\forall x \in I) \quad (L-\epsilon)g(x) < f(x) \le L h(x)$$ implies the quantified statement $$(\forall x \in I) \quad L g(x) \le f(x) \le L h(x).$$ Now apply the squeeze theorem.
A: Generally no: as a rule of thumb, limits only preserve nonstrict inequalities. Thus your second to last line is not always legitimate. In general you get the slightly weaker statement
$$L-\varepsilon \leq \lim_{x \to x_0} f(x) \leq L.$$
One can see this by looking at an example like this: although $1/n>0$ for all $n \in \mathbb{N}$, $\lim_{n \to \infty} 1/n = 0$.
In this particular case you actually do get the strict inequality, by instantiating the original statement for $\varepsilon/2$, deriving
$$L-\varepsilon/2 \leq \lim_{x \to x_0} f(x) \leq L$$
by the preceding procedure, and then noting that this implies that
$$L-\varepsilon < \lim_{x \to x_0} f(x) \leq L.$$
But working this way really does not do you any good relative to just working with the nonstrict inequality.
