Originally Posted by

**Deveno** You seem to be concerned about the "hypothesis of the squeeze theorem being satisfied". Many people actually get the definition of a limit incorrect. They think:

$\displaystyle \lim_{x \to a} f(x) = L$ means:

"For any $\epsilon > 0$ there exists a $\delta > 0$ such that $|x - a| < \delta \implies |f(x) - L| < \epsilon$".

This is absolutely wrong, the correct definition is:

"For any $\epsilon > 0$ there exists a $\delta > 0$ such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$".

The interval $(a -\delta,a) \cup (a,a+\delta)$ is what is intended for $x$ to lie in: this is also called a "deleted neighborhood" of $a$.

The formal statement of the squeeze theorem does not require $g(x) \leq f(x) \leq h(x)$ for ALL points in a neighborhood of $a$ ($a = 0$ in this case), but rather all points in a deleted neighborhood of $a$.

Note that if, in fact, $g(x) < f(x) < h(x)$ in said deleted neighborhood (that is, the inequality is strict) this implies the (looser) restriction $g(x) \leq f(x) \leq h(x)$.

Let's prove the simple statement: $p < q \implies p \leq q$ via truth-tables. First off, note that $p \leq q$ is shorthand for the compound statement:

$(p < q) \vee (p = q)$.

Now, if $p = q$ or $p > q$, the implication is true, no matter if the consequent is true or false (a false statement can imply anything). So, without loss of generality, assume $p < q$ is true. Then the implication is true if and only if the consequent is true.

Since the consequent is an "or" statement, it is true if either $p = q$ is true, or $p < q$. If $p < q$, then $p = q$ is false and $p < q$ is true. So our implication has the form:

T --> (F or T), equivalent to:

T --> T, equivalent to:

T

Your objection that $f$ continuous on $(a,b)$ does not mean $f$ is continuous on $[a,b]$ is totally spurious. It is OBVIOUS that $(a,b) \subseteq [a,b]$, and therefore if $x \in (a,b)$, then $x \in [a,b]$, by the DEFINITION OF SUBSET.

This does NOT mean $[a,b]$ has every property $(a,b)$ has, you have this exactly BACKWARDS: for example $b$ is in the set $[a,b]$, but it most assuredly NOT in $(a,b)$.

In short, ProveIt is ABSOLUTELY CORRECT. It is you who are confused, and I suggest you actually LEARN some mathematics before pontificating about it.