So there's a part of the proof of L'hopital rule given in Pugh's Real Mathematical Analysis that I don't understand. I was going to ask a question, but it turns out that another user had the same confusion, and asked a question earlier(Clarification of L'Hopital Proof Pugh), nobody answered it and it only has a comment, but the comment doesn't address the confusion of the user.

Just like this user, I don't understand why the constraints

$$\begin{align}|f(t)+g(t)| &< \frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)} \\ |g(t)| &< \frac{|g(x)|}{2}.\end{align}$$ imply that $$\frac{g(x)f(t)-f(x)g(t)}{g(x)[g(x)-g(t)]} < \epsilon/2$$

Is it fine if I start a new post?. The other option is to start a bounty, but I have little reputation and I don't want to lose some of my privileges.

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    $\begingroup$ Would you be agreeable for me to post a bounty on your behalf? I think one of the standard reasons will do, such as this Question needs a canonical answer (or words to that effect). $\endgroup$
    – hardmath
    Oct 18, 2019 at 1:44
  • $\begingroup$ That'd be great, I'd be very glad if you do :D $\endgroup$ Oct 18, 2019 at 14:27
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    $\begingroup$ The first constraint has a typo: it is $|f(t)|+|g(t)|$, not $|f(t)+g(t)|$. See the original excerpt here. $\endgroup$
    – user9464
    Oct 18, 2019 at 15:26
  • $\begingroup$ @DonlansDonlans: Have you had a chance to read the Answers that were posted on the previous Questions? If you have some feedback I'd be happy to take it into consideration in awarding the bounty. $\endgroup$
    – hardmath
    Oct 20, 2019 at 14:56
  • $\begingroup$ @hardmath Yes,I have read them, DanielWainfleet's answer and Jack's answer have been the most useful to me. Though if I had to choose, I'd go for Jack's answer, he explained why the constraints imply the inequality that I mentioned on this post,which was what I specifically didn't understand.Thank you very much for starting the bounty :D $\endgroup$ Oct 21, 2019 at 15:15
  • $\begingroup$ @DonlansDonlans: Thanks for the feedback. I'll consider it when awarding the bounty. I've read and upvoted all the answers posted before I asked you to check them out. $\endgroup$
    – hardmath
    Oct 21, 2019 at 15:17
  • $\begingroup$ @hardmath: Thanks for your kindness and the bounty. $\endgroup$
    – user9464
    Oct 23, 2019 at 18:02

1 Answer 1


I have awarded the bounty on the Question at issue, wishing that I could have done more than just upvote other Answers provided.

One thing that in hindsight would have been helpful is an Edit to that Question that pins down where in Pugh's book that proof is found. There seem to have been more than one editions/revisions, so a more complete citation (e.g. year of publication) might have helped.

Based on a Google Books preview I think the proof appeared around page 144 in the earliest (2002/2003) edition, while the excerpt shown in Jack's Answer to the old Question shows it at page 154.

  • 1
    $\begingroup$ Indeed, the excerpt in my answer is from the second edition. As I can see from the first edition of Pugh's book, the proof is the same in both editions and I find no typos in either one. $\endgroup$
    – user9464
    Oct 23, 2019 at 18:06

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