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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.

Stephen K.
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    Why do you think the statement is true? The definition of limit does not seem to imply it in some obvious manner. – Paramanand Singh Jul 19 '17 at 19:52
  • You are correct! I've been thinking about this over the past day and have indeed just come up with a counter example where the statement does not hold for a case where $f$ is not continuous on $A$. (See figure below, where $f$ is the union of $n\in\mathbb{N}$ lines, each with positive slope and defined on a left-open interval; therefore, there is no $c$ such that the statement holds even though the limit of the function goes to $\infty$ at 0). However, if the conditional to the statement is narrowed such that $f$ is required to be continuous on $A$, then I think the statement holds. – Stephen K. Jul 19 '17 at 21:00
  • Figure of counterexample here: https://preview.ibb.co/ec5qQQ/counterexample.jpg – Stephen K. Jul 19 '17 at 21:09
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    Continuity doesn't suffice. If you remove all the intervals $\bigl[\frac{1}{n}, \frac{1}{n} + 2^{-n}\bigr]$ from $A$, your counterexample still works and is then continuous on $A$. It is sufficient to demand that $f$ is continuous and that $\overline{A} \subset A \cup {a}$. – Daniel Fischer Jul 21 '17 at 12:31
  • Indeed, that appears to be the case! Can you clarify what you mean by $\bar{A}\subset A\cup{a}$? Alternatively, is it sufficient to demand that $f$ be continuous on $A$ and where $A$ is an interval? – Stephen K. Jul 21 '17 at 12:55
  • BTW I am unable to understand why we need this lemma to prove L'Hospital's Rule. As far as I am aware the proof is almost an immediate consequence of Cauchy's Mean Value Theorem. – Paramanand Singh Jul 21 '17 at 16:20
  • The particular proof of L'Hospital's Rule that my question is in reference to is from Bartle and Sherbert's Introduction to Real Analysis (see https://i.stack.imgur.com/8GCqZ.jpg), which makes reference to a prior proof that can be found here. In particular, the part that my question above is in reference to is the statement right after equation (6). – Stephen K. Jul 21 '17 at 17:53
  • @ParamanandSingh: and yes, the proof is indeed based on Cauchy's Mean Value Theorem. This particular lemma builds on that as seen in the link above. – Stephen K. Jul 21 '17 at 18:04
  • Still I don't get why you need your lemma for equation $(6)$. That equation is a trivial consequence of Cauchy's mean value theorem. If you want I can provide an explanation of the proof there in much simpler terms. But that could be discussed in another question and this post can be left dealing with your personal lemma. Also the proof in your book is simple but presented in a very complicated fashion with lot of symbols. You are perhaps lost because of all that ugly stuff. Most analysis textbooks go to great lengths of symbolism and formalism to make proofs boring. – Paramanand Singh Jul 21 '17 at 19:06
  • Sorry, my comment might have been confusing. The lemma is not needed for equation (6), rather the statement after equation 6: "we may assume that $0<g(c)/g(\alpha)<1$ for all $\alpha\in(a,c)$..." – Stephen K. Jul 21 '17 at 19:18
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    Oh that part is simple enough and follows from the fact that derivative $g'$ is negative. But the book should have mentioned it as it may not be obvious. Since $g'$ does not vanish, it is of constant sign by Darboux theorem. And if it were positive $g$ would not tend to $\infty$ as $x\to a^{+} $. – Paramanand Singh Jul 21 '17 at 19:40
  • I must say your confusion has a happy impact namely investigation and finally proof of a lemma. – Paramanand Singh Jul 21 '17 at 20:00

2 Answers2

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I believe I have a proof for the special case when the domain is an interval. It is interesting to me that the proof of this simple assertion is not so trivial (although if there is a simplier route to getting this result, I would be interested in finding out).

Theorem Let $f$ be a continuous function on interval $(a,b)\!\subset\!\mathbb{R}$. If $\displaystyle{\lim_{x\to a+}f(x)\!=\!\infty}$, then, given any $\alpha\!\in\!\mathbb{R}$, there exists $c\!\in\!(a,b)$ such that: (1) $\alpha\!\leq\!f(c)$, and (2) $x\!\in\!(a,c)$ implies $f(c)\!<\!f(x)$.

Proof From the definition of the infinite right-handed limit, given any $\alpha\!\in\!\mathbb{R}$, there exists $\delta\!>\!0$ (where we can assume that $a+\delta\!<\!b$) such that: $$\qquad\qquad\ 0\!<\!x\!-\!a\!<\!\delta\ \ \mbox{implies}\ \ \alpha\!<\!f(x).\qquad\qquad(*)$$ Since $a\!+\!\delta\!\in\!I$, then either: (1) $f(a\!+\!\delta)\!=\!\alpha$, (2) $f(a\!+\!\delta)\!<\!\alpha$, or (3) $f(a\!+\!\delta)\!>\!\alpha$.

  • Case (1), where $f(a\!+\!\delta)\!=\!\alpha$: Let $c\!\equiv\!a+\delta$ and the assertion is proved.
  • Case (2), where $f(a\!+\!\delta)\!<\!\alpha$: In addition to this case's assumption, one also has $f(a\!+\!\delta/2)\!>\!\alpha$ by the implication $(*)$ above. Therefore, since $f(a\!+\!\delta)<\alpha<f(a\!+\!\delta/2)$, by Bolzano's Intermediate Value Theorem, there exists a number $k\!\in\!(a\!+\!\frac{\delta}{2},a\!+\!\delta)$ such that $f(k)\!=\!\alpha$. A second application of Bolzano's theorem results in a number $l\!\in\!(k,a\!+\!\delta)$ such that $f(a\!+\!\delta)\!<\!f(l)\!<\!\alpha$. However, this contradicts implication $(*)$ above so that $f(a\!+\!\delta)\!\not<\!\alpha$ and this case is not possible.
  • Case (3), where $f(a\!+\!\delta)\!>\!\alpha$: Define $\alpha'\!\equiv\!f(a\!+\!\delta)$. Again, the infinite limit definition states that there exists $\delta'\!>\!0$ such that: $$\qquad\qquad\ 0\!<\!x\!-\!a\!<\!\delta'\ \ \mbox{implies}\ \ \alpha'\!<\!f(x).\qquad\qquad(**)$$ It must be that $\delta'\!\leq\!\delta$ (otherwise a contradiction would result for the point $a\!+\!\delta$ in (**) above). If $\delta'\!=\!\delta$, then let $c\!\equiv\!a+\delta'$ and the assertion is proved. If $\delta'\!<\!\delta$, then from the Extreme Value Theorem, $f$ has an absolute minimum $f_{min}$ on $[a\!+\!\delta',a\!+\!\delta]$. Define the following set: $$X\equiv\{\ x\ :\ x\!\in\![a\!+\!\delta',a\!+\!\delta]\ \mbox{and}\ f(x)\!=\!f_{min}\},$$ which includes all points in that subdomain that have $f_{min}$ as their image. The question next becomes whether the infimum of this set is a member of the set (i.e., whether the set has a minimum). The answer is that it must, because if it doesn't, then $f$ would be discontinuous at the infimum contradicting the assertion's hypothesis that $f$ be continuous. Let $c\!\equiv\!\inf X$ and the assertion is proved.
Stephen K.
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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 $S\subset \mathbb R.$

DanielWainfleet
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  • Thanks for your reply Daniel. However, it is not clear to me how $\lim_{c\to a+}\sup {f(x): x\in (a,c)\cap A}=\infty$ is equivalent to the statement I am trying to prove. Indeed, $\sup {f(x): x\in (a,c)\cap A}$ does not appear to even be clearly defined since, given any $c$, such a set is unbounded from above and therefore the supremum does not exist. – Stephen K. Jul 19 '17 at 12:12
  • I am using the convention that a set not bounded above has a $\sup$ of $\infty.$ It is often convenient in such contexts to extend $\mathbb R$ to $\mathbb R \cup {\infty}.$ – DanielWainfleet Jul 19 '17 at 16:39
  • Even in the context of the extended reals $\mathbb{R}^{#}=\mathbb{R}\cup{\pm\infty}$, the supremum of the set ${f(x):x\in(a,c)\cap A}$ will always equal $\infty$ by definition even before the limit of $c\to a+$ is taken. How is this related/equivalent to the statement to be proved? I appreciate your comments. – Stephen K. Jul 19 '17 at 17:49