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L'Hospital's Rule states that $$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

can be applied when:

(1) $f$, $g$ are differentiable;

(2) $g'(x) \neq 0$ for $x$ near $a$ (except possibly at $a$);

(3) $\displaystyle\lim_{x\to a}f(x) = 0 = \displaystyle\lim_{x\to a}g(x)$, or $\displaystyle\lim_{x\to a}f(x) = \pm\infty = \displaystyle\lim_{x\to a}g(x)$; and

(4) the limit on the RHS exists or it equals $\pm\infty$.

$\ $

Why is proviso (2) necessary?

Edit: This question was asked in the context of a first undergraduate calculus course. Thus, the domains of $f$ and $g$ can be assumed to include a subset of $\mathbb R$ for which $a$ is an accumulation point, that has at most two connected components. I apologise to Paramanand if this changes the question - I intended to communicate this by tagging the question calculus. I think that this setup means that the quotient $f'/g'$ will be defined on an interval near $a$ when the criteria (1)-(4) hold.

ahorn
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    For the case $x \to \infty$ see this paper by Boas: http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/counterexamples-to-lh-pitals-rule – Hans Lundmark Oct 20 '14 at 17:49
  • @HansLundmark, the case discussed is for $x \to \infty$ for simplicity only, a similar (but messier) proof apllies to a finite $x$. – vonbrand Jul 27 '15 at 23:55
  • @vonbrand: Yes, I know. – Hans Lundmark Jul 28 '15 at 07:25
  • Funnily, there is one more hypothesis that must be added, very similar to the one that you are talking about: (2') $g(x) \ne 0$ for $x$ near $a$ (except possibly at $a$) - for exactly the same reason for which hypothesis (2) is needed. – Alex M. May 24 '16 at 16:43
  • @AlexM.: Neither $(2)$ nor $(2')$ i.e. $g(x) \neq 0$ is required. See my answer. – Paramanand Singh May 25 '16 at 03:28
  • @ahorn: Concerning that bizarre event when you temporarily earned a number of reputation points (40 or 50, I guess), it seems that you have been the target of serial voting, i.e. from what I see somebody gave you 7 upvotes in the timespan of 28 seconds - clearly an abnormal voting pattern. This triggered the serial voting detection software which led to you automatically losing 45 of these points. – Alex M. May 26 '16 at 21:42
  • @AlexM. Yes, I realised the next day that the system corrected the upvotes and took my reputation back. – ahorn May 27 '16 at 10:36
  • Most of the debate surrounding your question is caused by the fact that you do not specify a domain of definition for $f$ and $g$. Your statement of l'Hospital's theorem is very similar to the one on Wikipedia. There, $f$ and $g$ are assumed to be defined on $I \setminus {a}$, with $I$ an open interval. Do you accept this domain of definition? Do you want it to be pathological, like $\Bbb Q$ or a Cantor set? I believe that editing your question to clarify this ambiguity could settle the issue once and for all. – Alex M. May 29 '16 at 13:46

2 Answers2

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I assume that we are using the following definition of limit:

Let $f$ be defined in a certain neighborhood of $a$ and not necessarily at $a$. A number $L$ is said to be the limit of $f$ as $x \to a$ (written as $\lim\limits_{x \to a}f(x) = L$) if for every $\epsilon > 0$ it is possible to choose a $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$.

Note that this definition presupposes that $f$ must be defined in a deleted neighborhood of $a$ before we can talk of $\lim_{x \to a}f(x)$. Many textbook authors relax this condition and only need that $a$ should be an accumulation point of the domain of $f$. Note that this relaxed condition does not help us in case of L'Hospital's Rule because the proof of this rule is based on the use of mean value theorem which requires that we are dealing with intervals (for which $a$ is an end-point) rather than dealing with arbitrary domains (which have $a$ as an accumulation point).

Once we have taken the definition of limit mentioned above it is obvious that condition $(4)$ of L'Hospital's Rule in question automatically implies condition $(2)$. The existence of $\lim_{x \to a}f'(x)/g'(x)$ presupposes that the expression $f'(x)/g'(x)$ is defined in a certain deleted neighborhood of $a$. And this means that $g'(x) \neq 0$ in that deleted neighborhood. So the second condition is redundant.

However it is a good idea to mention this condition because it is easy to check this instead of checking $(4)$. And if this fails we are guaranteed that LHospital won't work and we don't need to check the condition that $f'(x)/g'(x)$ tends to a limit.

Further note that if we are given that $\lim_{x \to a}g(x) = 0$ and the fact that $g'(x) \neq 0$ as $x \to a$ then it follows via mean value theorem that $g(x) \neq 0$ as $x \to a$ and hence it is possible to talk about limit of $f(x)/g(x)$. The proof for L'Hospital is now easily completed via Cauchy's Mean Value Theorem.


Update: OP wants to know (via comment to this answer) about the difference between two definitions of limit $\lim_{x \to a}f(x)$ which I have mentioned in my answer. In one definition we require $f$ to be defined in a deleted neighborhood of $a$ and in another definition we require $a$ to be an accumulation point of the domain of $f$. The difference is best explained via an example where these two definitions lead to two different conclusions.

Let $$f(x) = \frac{\sin(1/x)}{\sin(1/x)}$$ then it is clear that $f$ is defined for all values of $x$ except $x = 0$ or $x = 1/n\pi$ where $n$ is a non-zero integer. Moreover $f(x) = 1$ at points where $f$ is defined. Thus the domain of $f$ is $$A = \mathbb{R} - B$$ where $$B = \{0\}\cup\{1/n\pi\mid n\text{ is a non-zero integer}\}$$ and $f(x) = 1$ for all $x \in A$.

It can be easily seen that $0$ is an accumulation point of $A$ (this statement means that every deleted neighborhood of $0$ contains a member of $A$). Also observe that $0$ is an accumulation point of $B$.

At the same time we can see that the function $f$ is not defined in a deleted neighborhood of $0$ (because every such neighborhood contains points of $B$ also).

Thus according to my definition of limit it is not possible to even talk about $\lim_{x \to 0}f(x)$ whereas according to the relaxed definition we can talk about $\lim_{x \to 0}f(x)$ because $0$ is an accumulation point of domain $A$ of $f$. Moreover according to this definition we have $\lim_{x \to 0}f(x) = 1$.

For a beginner who is studying calculus for the first time (meaning age 15-16 years) it is best to avoid terms like accumulation point, Bolzano-Weierstrass Theorem, open set, closed set etc (which are part of a course in real-analysis or point set theory/topology) and instead rely on simple definition of limit where the functions involved are defined on certain intervals except for some exceptional points of the interval where the function is not defined. The power of concept of limit is due to the fact that it can describe the behavior of a function near those exceptional points where the function is not even defined.

In fact this is ingrained in the definition of limit that limit of a function at a point $a$ has nothing to do with its value at point $a$ and has everything to do with its values near point $a$. Therefore it is a pre-requisite that the function must be defined in some neighborhood of $a$ except possibly at $a$ (and we say this compactly that $f$ must be defined in a deleted neighborhood of $a$) before we can even even talk about the limit of $f$ as $x \to a$ (or to use a different phrase, limit of $f$ at point $a$).


Just to clarify we are talking about functions defined on subsets of $\mathbb{R}$ and hence there is no need to think of complex domains.

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    Why the downvote?? – Paramanand Singh May 25 '16 at 13:29
  • What is a 'deleted neighbourhood' as opposed to just a 'neighbourhood'? – ahorn May 27 '16 at 10:39
  • @ahorn: A neighborhood of a point $c$ is any open interval which contains $c$. If $I$ is a neighborhood of $c$ then the set $I_{d} = I - {c}$ consisting of all points of $I$ except $c$ is called a "deleted neighborhood of $c$". Thus if we delete/remove $c$ from a neighborhood of $c$ then we get a deleted neighborhood of $c$. – Paramanand Singh May 27 '16 at 10:42
  • What is the difference between '$a$ has a deleted neighbourhood in the domain' and '$a$ is an accumulation point of the domain', in this context? I assume we consider the domain to be real because I don't know how else a derivative could be defined (complex perhaps). – ahorn May 28 '16 at 19:27
  • @ahorn: I will provide a reply to your comment in an update to my answer as it is too long to fit in a comment. – Paramanand Singh May 29 '16 at 08:02
  • @ahorn: I have updated my answer with more details on the difference between $f$ defined in a deleted neighborhood of $a$ and $a$ being accumulation point of domain of $f$. Let me know if you have any further doubt. – Paramanand Singh May 29 '16 at 08:29
  • @AlexM.: Sorry but I could not get the intent of your comment. I was expecting that you would mention some lapses in my answer (because of which you find it unsatisfactory) but you instead want to say that I am doing "hair splitting". I don't see how i can really help to improve my answer to your satisfaction. – Paramanand Singh May 29 '16 at 09:06
  • Of course that $\lim \limits _{x \to 0} f(x)$ can be defined using deleted neighbourhoods! Remeber that the domain of $f$ in your example is $A$, not $\Bbb R$ (this is where you make your mistake), i.e. the function is $f :A \to \Bbb R$. Therefore, you must consider $A$ as a topological space itself, endowed with the induced subspace topology from $\Bbb R$. Then one may define $\lim \limits _{x \to 0} f$ on $A$ using deleted neighbourhoods (neighbourhoods in $A$, not in $\Bbb R$), and that is it. – Alex M. May 29 '16 at 09:11
  • @AlexM.: I think you are looking at this problem from the view of a person who had the knowledge of metric spaces, topological spaces and in general has done a course in real analysis. The current question does not need any of these things to answer it. It requires a basic understanding of limit, derivative. I have only tried to present the definition of limits of a function defined on subsets of $\mathbb{R}$ without using any notions of metric spaces/topological spaces. I think f I just wrote a one line answer that the second proviso is redundant it would have been better. Contd... – Paramanand Singh May 29 '16 at 09:17
  • @AlexM.: I added the remark about the "relaxed definition of limit concerning accumulation point" only to avoid such hair splitting. A definition of limit does not require any idea about limit points unless you specifically want to study limits in the context of metric spaces/topological spaces. – Paramanand Singh May 29 '16 at 09:18
  • If you adopt this point of view, then you are erasing one century of progress in thinking abot topological issues and you go back to the late 19th century. You may very well do it, but why? It's like avoiding electronic calculators just because you want to use an abacus. Feel free to do it, but is it reasonable? – Alex M. May 29 '16 at 09:27
  • @AlexM. Simpler is better. Everyone studying mathematics should aim for simplicity and economy. Powerful tools should be used to tackle powerful problems. But yes its my personal opinion! I finally wish to make it clear that we don't have any serious disagreement over this minor issue of definition of limit and whatever minor disagreement seems to exist is because of looking from two different viewpoints. It was nice interacting with you :) – Paramanand Singh May 29 '16 at 09:39
  • Part 1/2: @ParamanandSingh ever since the first version of your answer, you have used notions that a first-year student would not understand - i.e. a student who has not studied the $\epsilon - \delta$ definition of a limit and had exposure to metric spaces and topological spaces. I asked this question in 2014 when I was in first year and was first introduced to l'Hospital's rule, but now I am in third year and have experience of the abovementioned subjects. Thus, you are welcome to expand your ideas into metric spaces concepts. I think Alex has a good point in his... – ahorn May 29 '16 at 12:30
  • Part 2/2: ...comment above that $f$ is only well-defined on $A$. I would like it if you could provide a counter-argument. I'm leaning towards asking a new question based on the debate we are currently having about accumulation point versus deleted neighbourhood, just to get some fresh air on the matter. – ahorn May 29 '16 at 12:30
  • @ahorn: My answer specifically tries to avoid notions like accumulation point which are not part of vocabulary of a beginner in calculus. That's why I avoid it and assume the definition based on neighborhoods. The part about accumulation points was added for people who have attained some maturity about real-analysis topics and who might think my assumed definition of limit as too simplistic. BTW $\epsilon, \delta$ has to be part of vocabulary of any beginner in calculus who wants to understand calculus theorems rather than solve routine derivative/integral problems. contd... – Paramanand Singh May 29 '16 at 12:37
  • @ahorn: I don't understand what "counter argument" you are expecting. In my update i have a function defined on a set $A$ which is not exactly an interval and. Also the function is such that $0$ is an accumulation point of $A$ as well as its complement $B$. – Paramanand Singh May 29 '16 at 12:39
  • @ahorn: To be frank I am not sure if you have really understood the crux of my answer here. The crux is that second proviso is unnecessary. whatever definition of limits you wish to chose does not impact the redundancy of proviso $(2)$. – Paramanand Singh May 29 '16 at 12:42
  • @AlexM. w.r.t. your comment above, I think what Paramanand was trying to say is that $f$ is not defined on a (real) interval around $a$ - an interval is connected by definition - and the MVT requires intervals. I wonder if the proof of l'Hospital's always requires an interval around the limit point. – ahorn May 29 '16 at 13:09
  • @ahorn: This is exactly what i was saying. There is no proof (known to me) of L'Hospital without use of mean value theorems. – Paramanand Singh May 29 '16 at 13:22
  • @ParamanandSingh I think that the necessary condition of an interval around $a$ also rests on proviso (1), which implies that $f'/g'$ is continuous around $a$ when (2) is satisfied. (edited) – ahorn May 29 '16 at 13:34
  • @ahorn: from $(4)$, $f'/g'$ exists near $a$. It is not guaranteed to be continuous. – Paramanand Singh May 29 '16 at 13:37
  • Good answer +1. Have a look at this https://math.stackexchange.com/a/2379732/148510 – RRL Aug 02 '17 at 08:25
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Remember that in order to be able to speak about the limit of some function $h : D \to \Bbb R$ at a point $a$ (with $D$ an open subset), you need $a$ to be an accumulation point for $D$ (a limit point of $D$, as it is also called) - in particular, this means that $h$ must be defined in some neighbourhood $a$ (save for, possibly, in $a$ itself).

Using the above and letting $h = \frac {f'} {g'}$, you want $h$ to be defined on some neighbourhood of $a$, and this amounts to $h$ having a non-zero denominator in some neighbourhood of $a$, i.e. $g'(x) \ne 0$ for $x$ in a neighbourhood of $a$ (possibly with $a$ omitted - this one being often the case, in practice).

To conclude, the condition on the denominator has to do only with having the fraction well-defined (i.e. mathematically meaningful).

Alex M.
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    IIRC, you need an open interval around $a$ to have $g'$ nonzero because you need to apply the MVT or generalized MVT and need $f/g$ or $f'/g'$ or something like it well defined on an interval. This is impossible with a function involving $\sin(1/x)$ and in fact, I believe you can come up with an appropriate function where lhospitals rule fails. – abnry May 25 '16 at 01:08
  • I remember discovering in my analysis book that this precise condition was misstated (Abbott's understanding analysis). I may be able to dig up my notes. – abnry May 25 '16 at 01:11
  • @MathematicsStudent1122: The definition of a limit as $x \to a$ where we only assume that $a$ is an accumulation point of domain of the function is fine. But it does not help when we are dealing with L'Hospital rule. The proof of LH rule critically requires the use of mean value theorems which hold on intervals and then it appears reasonable to use the simpler definition of limit (which is taught well before a student hears about accumulation point). – Paramanand Singh May 25 '16 at 03:21
  • @ahorn: Isolated points are not accumulation points. This is why I find Paramanand Singh's answer not satisfactory. – Alex M. May 27 '16 at 11:26
  • @AlexM. You're right. I got the definition of an accumulation point wrong. My metric spaces textbook says that it must be zero distance from the rest of the set. – ahorn May 28 '16 at 09:02
  • I'm accepting this answer because when you talk about $D$, it is a general case scenario. After the first paragraph, the specific context of this question is employed. I have yet to be convinced that there is a difference between the existence of a neighbourhood of a point and that point being an accumulation point, when we are in $\mathbb R$. – ahorn May 28 '16 at 20:19
  • @AlexM.: I have never said that isolated points are accumulation points in my answer. Please let me know in somewhat detail as to why you find my answer unsatisfactory. I would like to improve my answer in whatever possible way I can. – Paramanand Singh May 29 '16 at 08:36
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    BTW if $a$ is an accumulation point of domain $D$ of $h$ then it does not mean that $h$ must be defined in a neighborhood of $a$ except possibly at $a$. You are trying to mix the two definitions of limit (of my answer) and trying to imply that they are same. They are not same. One with accumulation point stuff is more general but not applicable to the case of L'Hospital. This is what I have tried to state very clearly and explicitly but it appears that I have been misunderstood somehow. – Paramanand Singh May 29 '16 at 08:45
  • @ParamanandSingh Sorry for the misunderstanding about isolated points - I deleted a comment. W.r.t. your last comment, I have opened up a new question. – ahorn May 29 '16 at 13:22