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The following is taken from the 2nd Edition of Stephen Abbott's book "Understanding Analysis". I must admit that I am a huge fan of this book.

Theorem 5.3.6 (L'Hopital's Rule, 0/0 case) Let $f$ and $g$ be continuous on an interval containing $a$, and assume $f$ and $g$ are differentiable on this interval with the possible exception of the point $a$. If $f(a) = g(a) = 0$ and $g'(x) \neq 0$ for all $x \neq a$, then $ \lim_{x \to a} \frac{f'(x)}{g'(x)} = L $ implies $ \lim_{x \to a} \frac{f(x)}{g(x)} = L. $

Exercise 5.3.10 Let $f(x) = x \sin(1/x^4)e^{-1/x^2}$ and $g(x) = e^{1/x^2}$. Using the familiar properties of these functions, compute the limit as $x$ approaches $0$ of $f(x)$, $g(x)$, $f(x)/g(x)$ and $f'(x)/g'(x)$. Explain why the results are surprising, but not in conflict with the content of Theorem 5.3.6.

I found these limits to be $\lim_{x \to 0} f(x) = 0$, $\lim_{x \to 0} g(x) = + \infty$, $\lim_{x \to 0} f(x)/g(x) = 0$ and $\lim_{x \to 0} f'(x)/g'(x) = 0$.

Theorem 5.3.6 does not apply directly, since none of these functions or combinations of functions is even defined at $0$. Nor can $g(x)$, for example, be continuously extended to $0$.

There is a footnote to the exercise saying "A large class of "counterexamples" of this sort to L'Hopital's Rule are explored in [4]." Unfortunately I am away from my copy of the book and am unable to see what is reference [4].

My question is - in what sense are these results surprising, or like "counterexamples" to L'Hopital's Rule ?

Simon
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  • It's probably a typo and should have been $g(x) = e^{-1/x^2}$. – Daniel Fischer Dec 12 '16 at 12:56
  • @DanielFischer Nop, it appears exactly as the OP wrote it...Second edition. – DonAntonio Dec 12 '16 at 12:59
  • @DonAntonio I mean it's probably a typo in the book. – Daniel Fischer Dec 12 '16 at 13:00
  • I have the same textbook, second edition. My copy has $g(x)=e^{-\frac{1}{x^2}}$ for Exercise 5.3.10 on page 162. –  Nov 15 '20 at 16:42
  • @SprNtndoChlmrs, I guess either a typo was corrected in a reprinting of the 2nd Edition (is that how printing works ?) or I might have been referring to an electronic issue which might have differed from my print issue by a typo, or I was mistaken and there was no typo. My print copy of the book is packed away for now in boxes, and I think my right to access an electronic version has expired, so I won't check it right now. – Simon Nov 15 '20 at 18:41

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Everything makes sense if we assume there's a typo in the book, and it was intended to have $g(x) = e^{-1/x^2}$. Then we have $f(x) \to 0$, $g(x) \to 0$ and

$$\frac{f(x)}{g(x)} = x\sin (x^{-4}) \to 0$$

as $x \to 0$, but

\begin{align} \frac{f'(x)}{g'(x)} &= \frac{\sin (x^{-4})g(x) -4x^{-4}\cos(x^{-4})g(x) + x\sin (x^{-4}) \frac{2}{x^3}g(x)}{\frac{2}{x^3}g(x)} \\ &= \frac{x^3}{2}\sin (x^{-4}) - \frac{2}{x}\cos (x^{-4}) + x\sin (x^{-4}) \end{align}

doesn't have a limit at $0$, due to the

$$-\frac{2}{x}\cos (x^{-4})$$

term. The other two terms tend to $0$.

This is "surprising" in so far as $\lim\limits_{x\to 0} \dfrac{f(x)}{g(x)}$ is not the same as $\lim\limits_{x\to 0} \dfrac{f'(x)}{g'(x)}$ - the latter doesn't exist, the former does. It's not a contradiction to theorem 5.3.6, since part of the hypothesis of that theorem is that $\lim\limits_{x\to 0} \dfrac{f'(x)}{g'(x)}$ exists. Since this is not the case here, the theorem doesn't say anything about the behaviour of $\dfrac{f(x)}{g(x)}$.

Daniel Fischer
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  • Thank you very much Daniel. That makes sense. If instead we assume that $f(x) = x \sin (1/x^4) e^{1/x^2}$ and $g(x) = e^{1/x^2}$, then we get again that the limit of the ratio of the derivatives does not exist, while the limit of the ratio of the functions does exist. But I think the typo you suggest is a more likely one ! – Simon Dec 12 '16 at 13:21
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    Yes, if we look at $x\sin (x^{-4})e^{1/x^2}$, that doesn't have a limit at $0$. In the context, we expect that both functions have limit $0$. – Daniel Fischer Dec 12 '16 at 13:23
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    Yes you are right. Assuming the typo is corrected per your suggestion, then by continuously extending $f$ and $g$ to $0$, we can verify all the hypothesis of the theorem, except for the existence of the limit of the ratio of the derivatives. – Simon Dec 12 '16 at 13:27