Lemma used in Proof of L'Hôpital's Rule for Indeterminate Types of $\infty/\infty$ This question arises from an unproved assumption made in a proof of L'Hôpital's Rule for Indeterminate Types of $\infty/\infty$ from a Real Analysis textbook I am using. The result is intuitively simple to understand, but I am having trouble formulating a rigorous proof based on limit properties of functions and/or sequences.
Statement/Lemma to be Proved:
Let $f$ be a continuous function on interval $(a,b)\!\subset\!\mathbb{R}$. If $\displaystyle{\lim_{x\rightarrow a+}\!f(x)\!=\!\infty}$, then, given any $\alpha\!\in\!\mathbb{R}$, there exists $c\!>\!a$ such that $x\!\in\!A\cap(a,c)$ implies $\alpha\leq f(c)<f(x)$.
Relationship with Infinite Limit Definition:
At first glance, this may appear to slimply be the definition of right-hand infinite limits:


*

*$\displaystyle{\lim_{x\rightarrow a+}\!f(x)\!=\!\infty}$ is defined to mean: given any $\alpha\!\in\!\mathbb{R}$, there exists $\delta\!>\!0$, such that $x\!\in\!A\cap(a,a+\delta)$ implies $\alpha<f(x)$.


However, the main difference is that the result I am interested in forces an association between the "$\delta$" and the "$\alpha$" (where $\alpha=f(a+\delta)$, i.e., it forces $c\!\equiv\!a+\delta$ to be in the domain of $f$).
EDIT: The statement to be proved that I originally presented did not require $f$ to be continuous on $A$. However, this was added to address the comment and counterexample below.
EDIT #2: Again, a helpful user (@DanielFischer) commented that the statement after the first edit needed yet an additional limitation--i.e., that $A$ must also be an interval--for it to hold.

Assuming that $a$ is a lower limit point of $A$ \ $\{a\},$ the statement $\lim_{x\to a+}f(x)=\infty$ is equivalent to $$\lim_{c\to a+}\inf  \{f(x): x\in (a,c)\cap A\}=\infty.$$ And the statement you wish to derive from $\lim_{x\to a+}f(x)=\infty$ is equivalent to $$\lim_{c\to a+}\sup \{f(x): x\in (a,c)\cap A\}=\infty.$$ Which is obvious because $\sup S\geq \inf S$ for any $S\subset \mathbb R.$ 
