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I am familiar with the following version of L'Hôpital's rule: Let $f,g:I\to\Bbb R$ be differentiable on the interval $I$, further assume that $g'(x)\ne0$. Let $a\in I$ and assume the limit $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ exists. If it is the case that $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$ or $\lim_{x\to a}g(x)=\infty$ then the limit $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists, and it coincides with $\lim_{x\to a}\frac{f'(x)}{g'(x)} $. Now my question is, why do such an overwhelming number of books state the theorem with the requirement that $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$ or $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty$? In the latter case, $\lim_{x\to a}f(x)=\infty$ is really not needed. But I have seen the theorem stated like this even in non introductory books. What's the deal? thanks

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Your question is really a pedagogical question, not an analysis question.

First, you're right: when $g(x) \rightarrow \infty$ the condition $f(x) \rightarrow \infty$ is not needed: a statement of the result in this form can be found in Rudin's Principles of Mathematical Analysis. So why is this condition included in most elementary treatments, e.g. virtually all freshman calculus books?

As a veteran instructor of freshman calculus, I can say that L'Hopital's Rule is such a source of confusion and pitfalls to students that I almost wish we didn't cover it at all. The statement is simply too subtle, and students often misuse it. (Congratulations by the way on stating it correctly!)

We generally present the rule in the context of a larger discussion of Indeterminate Forms, i.e., limits like $0/0$ and $\infty/\infty$ which cannot be evaluated from the above form but depend on the actual functions involved. In freshman calculus it is relatively rare that $\lim_{x \rightarrow a} f(x)$ simply does not exist, and if the limit is some finite number $L$ then $L/\infty$ is not an indeterminate form: it is always $0$. Thus in practice one would not use this extension of the rule in freshman calculus, and including it would run the serious risk of further confusing students on the concept of an indeterminate form.

(Nevertheless I state the rule this way in my honors calculus notes.)

It would be interesting to see as a rebuttal, an example of a limit that can be computed by this "freshman excluded" case of L'Hopital and cannot otherwise be easily computed by the methods of freshman calculus. Can anyone think of a natural example, one that could conceivably be shown to freshman calculus students?

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Maybe something like $\lim_{x\rightarrow \infty}{\frac{x\sin{x}}{1+x^3}}$? The limit $\lim_{x\rightarrow \infty}{x\sin{x}}$ doesn't exist, but after l'Hospital we have $\lim_{x\rightarrow \infty}{\frac{\sin{x}+x\cos{x}}{3x^2}}=\lim_{x\rightarrow \infty}({\frac{\sin{x}}{3x^2}+\frac{\cos{x}}{3x}})$ which is $0$. – Ludolila Feb 16 '13 at 20:21
Thought now I see that is can be easily computed using other (freshman) methods :) – Ludolila Feb 16 '13 at 20:25

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