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We've got three different phrasings of L'Hôpital's rule and got a little bit confused about the subtle differences between them.


A

Let $-\infty \leq a < b \leq +\infty$ and let $f,g:]a,b[ \to \mathbb{R}$ be two functions, differentiable on $]a,b[$, and let $g'(x) \not= 0 \forall x \in ]a,b[$.

If further more on of the following cases applies

Case 1. $\lim\limits_{x \to a^+} f(x) = \lim\limits_{x \to a^+} g(x) = 0$.

Case 2. $\lim\limits_{x \to a^+} g(x) = \pm \infty$.

and $L := \lim\limits_{x \to a^+} \frac{f'(x)}{g'(x)}$ exists,

then we have $\lim\limits_{x \to a^+} \frac{f(x)}{g(x)} = L$.


B

Let $-\infty \leq a < b \leq +\infty$ and let $f,g:]a,b[ \to \mathbb{R}$ be two functions, differentiable on $]a,b[$, and let $g'(x) \not= 0 \forall x \in [a,b]$.

If further more on of the following cases applies

Case 1. $\lim\limits_{x \to a^+} f(x) = \lim\limits_{x \to a^+} g(x) = 0$.

Case 2. $\lim\limits_{x \to a^+} f(x) = \lim\limits_{x \to a^+} g(x) = \pm \infty$.

and $L := \lim\limits_{x \to a^+} \frac{f'(x)}{g'(x)}$ exists,

then we have $\lim\limits_{x \to a^+} \frac{f(x)}{g(x)} = L$.


C

Let $a,b \in \mathbb{R}$ with $a < b$ and let $f,g:[a,b] \to \mathbb{R}$ be two functions, differentiable on $[a,b]$, and let $g'(x) \not= 0 \forall x \in [a,b]$.

If further more $f(a) = g(a) = 0$ and $L := \lim\limits_{x \to a^+} \frac{f'(x)}{g'(x)}$ exists,then we have $g(x) \not= 0 \forall x \in ]a,b]$ and $\lim\limits_{x \to a^+} \frac{f(x)}{g(x)} = L$.


A and B seem to be nearly the same

  • A is lacking the condition for $\lim f(x)$ in case 2 - isn't this necessary?
  • B requires $g'(x) \not= 0$ on the closed set $[a,b]$ - why is this? $g$ itself isn't even defined on $[a,b]$ and actually we're not working on the extended reals so the following leads to trouble $g'(x) \not= 0 \forall x \in [-\infty,+\infty]$.

C is somehow different

  • Since $a$ isn't allowed to be $-\infty$ and $b$ is not allowed to be $+\infty$, C is not including the same functions as A and B.
  • It requires differentiability and $g'(x) \not= 0$ on the closed set $[a,b]$ in contrast to A and B.
  • It lacks the second case. So C only helps us, if we can't calculate the limit of $\frac{f(x)}{g(x)}$ since $f(x)=0$ and $g(x)=0$.
  • We additionally get information about $g(x)$ on the half-open interval $]a,b]$.

Maybe someone can help explaining the subtle differences and how the actually come into account.

Further more I noticed, that the rule is considering $\lim\limits_{x \to a+}$ - when applying this rule we often just write $\lim\limits_{x \to a}$. This is just being not completely strict?

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Noone who can help? –  meinzlein Mar 25 '12 at 15:43
    
About the case A.2, see maa.org/programs/faculty-and-departments/… (read the post-proof commentary). –  Martín-Blas Pérez Pinilla Jan 30 at 11:20
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1 Answer

up vote 2 down vote accepted

Well, you seem to have understood everything correctly already...

(A) is the version given in Rudin's Principles. It's true that the condition on $f$ in (B) in the case when $g \to \pm\infty$ isn't needed, and also that it seems very strange to require something from $g'(a)$ in (B). I would say forget about (B) and go with (A).

The extra information about $g$ gained in (C) doesn't seem so interesting that I would find it worth formulating a special rule for that, but who knows, maybe it's useful sometimes?

And regarding $x \to a^+$: often $f/g$ is defined on a punctured interval around $a$, and one really wants to investigate the limit as $x \to a$. This can be done by invoking the rule for $x \to a^+$ separately and for $x \to a^-$ separately (considering then some interval $c < x < a$). Since both investigations often involve exactly the same calculations, one can simply do the two cases together and write "$x \to a$".

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