51

Upon looking at yet another question on this site on evaluating a limit explicitly without l'Hopital's rule, I remembered that one of my professors once said something to the effect that in Europe (where he is from) l'Hopital's rule isn't "overused" like it is here in the USA.

My question is, is there some reason not to use l'Hopital's rule when you have an indeterminate form? I know other techniques but l'Hopital is certainly my go-to. Is there some reason for hostility toward l'Hopital's rule?

nbubis
  • 31,733
  • 7
  • 76
  • 133
user241336
  • 511
  • 5
  • 3
  • 3
    It may have to do with a view to what the point of teaching mathematics in secondary school is. If you believe that one of the main reasons to teach mathematics is for problem solving skills, then I agree that l'Hopital's rule removes, in a sense, the problem-solving aspect of computing limits using other techniques (like squeeze theorem, manipulating the function's form like one would do with functions involving radicals, etc). – Hayden May 17 '15 at 18:33
  • 4
    Maybe because some times, using l'Hopital is overcomplication – there may be more elegant ways. – wythagoras May 17 '15 at 18:35
  • 7
    Duplicate of http://meta.math.stackexchange.com/questions/13008/why-are-people-so-interested-in-finding-limits-without-lh%C3%B4pitals-rule – user26486 May 17 '15 at 18:37
  • When taught to "see" how to evaluate a limit without l'Hospital one can see if a limit is a l'Hospitalble or not without need to take the derivatives explicitly. With lower chance of error while taking derivatives, too. @user241336 – Alexey Burdin May 17 '15 at 18:44
  • 15
    What is wrong is that it encourages students to not get to grips with the function, to concentrate on the tool and not the problem. – André Nicolas May 17 '15 at 18:55
  • For most functions defined by an analytical expression it's normally easier to apply l'Hôspital's rule. A similar situation occurs with some problems, which can be solved by establishing an algebraic equation or system of equations and solving it, but may also be solved without using equations at all. Just performing the appropriate arithmetic operations on the given data. This latter method requires relatively clever reasoning, while the algebraic method is more mechanic. – Américo Tavares May 18 '15 at 00:19
  • 1
    There is nothing wrong with the use of a powerful tool to solved a complex problem. But at the same time it does not seem wise to use a powerful tool for simple problems. Especially when a beginner is learning limits and continuity stuff L'Hospital's rule should be avoided at any cost. Its use right in the start of learning calculus does create severe impairment in understanding of limit concept. – Paramanand Singh May 18 '15 at 05:08
  • 2
    Is it possible it's just a fair warning for two reasons: (a) sometimes people tend to forget to check some conditions that must hold; and (b) sometimes L'Hopital's rule tends to blow up the function making it only harder to find the limit. – Willem Van Onsem May 18 '15 at 16:03
  • [Another related meta question](http://meta.math.stackexchange.com/questions/18857/using-taylor-expansion-on-a-limit-tagged-without-lhospital). – Hans Lundmark May 20 '15 at 09:22

7 Answers7

34

I can only guess that your professor feels it is an overpowered tool. Consider the limit $$\lim_{x\to 0} \frac{\sin(x)}{\sin(x)+\tan(x)}$$ There is a lot to be gained for a math student by spending more time on this limit. After all, the numerator is approaching zero, so it seems that the limit could be zero. The denominator is also approaching $0$ so maybe the limit doesn't exist. Clarity comes when you multiply through by $\frac{\csc(x)}{\csc(x)}$ The limit is $1/2$, which is actually sort of curious! (to me at least when I first learned it) It demonstrates a peculiarity of limits in that they can defy intuition. Mathematicians should experience this regularly. You might lose this sort of thinking experience if you just apply L'Hospital's rule off the bat. There are also limits like $$\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n$$ that as a first year calculus student I would not be able to do without L'Hospital's rule. Even using L'Hospital's rule I would have found that to be a tricky limit. I'm guessing your professor ultimately feels that L'Hospital's rule should be used as a last resort, when calculating the limit through other means is just not realistic (at that students current level of ability). Otherwise, the student should really spend time contemplating the limit and try to coax out the answer with some critical thinking. I know a number of my college professors felt this way about students using Tabular integration when doing integration by parts. I was a huge fan of Tabular integration because it made integrals like $$\int x^4\sin(x)\text{d}x$$ a breeze. Eventually I had a professor intentionally write all tests and quizzes so that Tabular integration wouldn't apply. This forced me to get a lot better at integration by parts, and that skill was truly a blessing when I later started learning about Fourier Series. If you continue on in mathematics you will get to real analysis. You will do limits much more rigorously, and L'Hospital's rule will probably not be available for use. The intuition you gain now from tangling with limits will benefit you down the road.

graydad
  • 13,684
  • 10
  • 24
  • 38
  • 10
    Is tabular integration a popular thing? I just looked it up because I didn't recognise the term, seems i've managed to slip past it. It does look like quite a neat trick though I have to admit. – Rammus May 17 '15 at 18:58
  • 2
    @Rammus I don't think it is extremely popular. I wasn't aware of any other undergrad students in my year that knew of it. I think my first calc teacher broke some sort of covenant when he taught my class ;) My college profs asked me not to teach it to anyone else until semester ended. – graydad May 17 '15 at 19:03
  • 24
    With $\lim_{x\to 0} \frac{\sin(x)}{\sin(x)+\tan(x)}$, in my head I just simplify it to $\lim_{x\to 0} \frac{x}{x + x}$ because differentiable functions are locally linear. – bjb568 May 17 '15 at 19:16
  • @bjb568 that is not how I thought when I first began calculus, and I imagine many other beginners would say the same – graydad May 17 '15 at 19:17
  • 2
    @graydad Now I'm curious. How could one use - even state - L'Hospital's rule before getting to derivatives? – Daniel Fischer May 17 '15 at 19:28
  • 2
    @DanielFischer you most certainly couldn't; a student would first need to spend some time with limits, then derivatives before learning L'Hospital's rule. I guess the problem I see in that, which I think I have struggled to relay in my answer, is that once you have such a useful tool as L'Hospital's rule, one may be inclined to stop critical thinking and just go into "derivative calculator" mode. – graydad May 17 '15 at 19:32
  • How would you even use L'Hospital for $\lim_{n\to\infty}(1+x/n)^n$? It would need to make a whole new transcendental function appear out of the blue, where the original limit has only rational operations ... – hmakholm left over Monica May 18 '15 at 01:47
  • @HenningMakholm you gotta twiddle it a little bit. You can rewrite it as $$\lim_{n \to \infty} e^{\frac{\ln\left(1+\frac{x}{n}\right)}{n^{-1}}}$$ – graydad May 18 '15 at 02:18
  • Isn't tabular integration merely integration by parts applied quickly to $x^n*f(x)$ without getting into the middle steps? – Gappy Hilmore May 18 '15 at 09:02
  • Tabular integration is a movie star. It appeared in the movie _Stand And Deliver_. – Steven Alexis Gregory May 18 '15 at 13:06
  • @grdgfgr: tabular integration is just doing integration by parts "in a table". One column contains the derivatives of $u$ and the other contains the integrands of $v$. Then you apply a mechanical algorithm. There's still the trickiness of picking which $u$ and $v$, which is the only hard part of integration by parts. – nomen May 18 '15 at 19:47
  • Just a quick note for the tabular integration: $$f(t)=\int\sin(xt)\ dx=-\frac{\cos(xt)}t+c_1\\\implies f^{(k)}(1)=\int x^k\sin(x)\ dx=\frac{d^k}{dt^k}\frac{\cos(xt)}t\bigg|_{t=1}+c_2$$ – Simply Beautiful Art Feb 19 '17 at 01:38
22

Students love L'Hopital because it's a magic wand they can wave over a problem, leaving intuition and actual thought undisturbed. A lot of students, at least in the U.S., don't understand the proof of the rule (it's not a trivial proof), and indeed don't really understand the definition of the derivative. Many problems that can be zapped by the magic wand are actually derivatives in disguise, and it's a shame if L'Hopital is invoked for these, since the student loses the opportunity to reconnect with one of the most important definitions in mathematics. My opinion: It's even overused on MSE (although I'm a fairly new member, I've seen a lot of it here). Example: We see things like $\lim_{n\to \infty} (\ln n)/n^2 = 0$ proved by L'Hopital on these boards. There's something wrong with that, again in my opinion, and I do thank you for reading my rant.

zhw.
  • 102,095
  • 6
  • 47
  • 106
  • I agree with your point on overuse of LHR or MSE. +1 for using the term "magic wand". – Paramanand Singh May 18 '15 at 05:01
  • How would you instead go about proving the equality in your example? I think L'Hôpital's rule is by far the simplest way in this case. Is there much more insightful way to go instead? – Ruslan May 20 '15 at 07:03
  • +1 for ...leaving intuition and actual thought undisturbed. – Fred Kline May 22 '15 at 01:34
  • If $b>1,$ then $n/b^n \to 0.$ Proof: Write $b=1+a, a> 0.$ Then by the (simple) binomial theorem, $b^n = (1+a)^n \ge 1 + na +[n(n-1)/2]a^2$ for $n>1.$ This gives the result, also giving us the intuition that we could carry out the binomial to much higher orders, showing actually that $n^p/b^n \to 0$ for any $p> 0.$ That $(\ln n)/n \to 0$ follows from this, by letting $n=2^m$ say. – zhw. May 22 '15 at 03:02
19

L'hopital's rule fails sometimes in the case of this example taken from [1.]; it cannot be used to evaluate the limit of \begin{align} \lim \limits_{x\rightarrow\infty} \frac{x}{\sqrt{x^2+1}}. \end{align}After applying L'hopital's once we get $$ \lim \limits_{x\rightarrow\infty} \frac{\sqrt{x^2+1}}{x}, $$ and then applying it again yields$$ \lim \limits_{x\rightarrow\infty} \frac{x}{\sqrt{x^2+1}}, $$ which we see will just indefinitely loop. I also agree with SalmonKiller that L'hopital's rule seems to just be an automated response by many; this is both a testament to its usefulness and a major pitfall, as there are times for which L'hopital's rule is not efficient.

[1.] Weisstein, Eric W. "L'Hospital's Rule." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/LHospitalsRule.html

Rammus
  • 2,311
  • 1
  • 10
  • 17
  • 8
    A committed L'H. Rule user could square the expression and use the Rule twice. – André Nicolas May 17 '15 at 18:53
  • 8
    @AndréNicolas Good point, but a L'H robot wouldn't :) – Rammus May 17 '15 at 18:55
  • 26
    $x\mapsto x/\sqrt{x^2+1}$ is increasing for positive $x$, and is also bounded above by $1$. Thus, the limit exists. Call it $L$. Then your first application of L'Hospital's rule says $L=1/L$, so $L=1$ ($L=-1$ can be excluded, since the function is non-negative). OK, I admit that this might not be the best way of doing it... :) – mickep May 17 '15 at 20:08
  • Whoa this is cool! How did you find this example? – Faraz Masroor May 17 '15 at 20:34
  • @FarazMasroor Check the reference in my post ^^. – Rammus May 18 '15 at 06:01
  • 10
    @mickep That is professional level L'Hing! – Rammus May 18 '15 at 06:04
  • 3
    @mickep oh such a complicated way... you at least need to find out that it's increasing. Why not just divide numerator and denominator by $x$ and use the usual properties of addition and division of limits? :) – Ruslan May 18 '15 at 15:55
  • 1
    @Ruslan I admit it was a bit of cheating not to show that the function is increasing. Here it goes: If $y>x>0$ then $y^2>x^2$, and $$y/\sqrt{y^2+1}-x/\sqrt{x^2+1}=\frac{y^2-x^2}{\sqrt{x^2+1}\sqrt{y^2+1}\bigl(y \sqrt{x^2+1}+x\sqrt{y^2+1}\bigr)}>0.$$ Of course I don't think anyone should use L'Hospital in this case! I'm just fooling around... – mickep May 18 '15 at 16:53
4

L'Hopital's is often described as "overused" because, students especially, often forget to check the conditions for L'Hopital's Rule. For example, if after using L'Hopital's the limit does not exist, then that says nothing about the original limit. L'Hopital's is often used before checking for other forms. E.g. if we try to evaluate a variation of the limit $$\lim_{x\to 0} \frac{\sin x}{x}$$ most students will use L'Hopital's, will make the process harder and longer. That's why many professors, especially oldschool ones, do not approve of using L'Hopital's rule too often and will often take off points for using it when it is uneccessary.

SalmonKiller
  • 2,030
  • 1
  • 14
  • 30
  • 18
    Using L'Hopital on $\lim_{x\to 0}\frac{\sin x}{x}$ is circular, since the proof of $\sin'(x)=\cos x$ usually uses that limit. – user26486 May 17 '15 at 18:40
  • @user31415 please expand. You mean in the derivative definition of the limit? – SalmonKiller May 17 '15 at 18:40
  • @wythagoras You have to calculate this limit to derive $\sin' = \cos$ in the first place, so actually using it here is quite circular. The proof I have usually seen uses the geometry to prove that $\sin(x) \leq x \leq \tan(x)$ for $0 \leq x < \pi/2$ and then uses the squeeze theorem. – Ian May 17 '15 at 18:41
  • @Ian You can derive $\sin' = \cos$ form $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, and $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$, can't you? Then you just need that the derivative of $e^x$ is $e^x$. – wythagoras May 17 '15 at 18:46
  • Or you can just use Taylor series. – wythagoras May 17 '15 at 18:47
  • 2
    @wythagoras That relies on a different definition of $\sin$ and $\cos$, which one should then check actually corresponds to the more familiar geometric definitions. The complex version also requires you to really be sure that differentiation of complex functions makes sense. – Ian May 17 '15 at 18:48
  • @Ian Euler's formula, which is not hard to prove. – wythagoras May 17 '15 at 18:50
  • 5
    @wythagoras It's not hard to prove that the series agree, provided we have established how infinite series work. But a proof that the series definition of $\sin$ and $\cos$ matches up with the geometric definition is at least as difficult as the geometric proof that $\sin'=\cos,\cos'=-\sin$, in my opinion. – Ian May 17 '15 at 18:51
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/23866/discussion-between-wythagoras-and-ian). – wythagoras May 17 '15 at 18:52
1

L’Hopital’s rule holds when the limit in contention satisfies four necessary conditions, else it is not applicable.

  1. $f$ and $g$ should be differentiable in an open interval $I$ containing the point $a$ for which we want to calculate the limit (except possibly at $a$). So, apply L’Hopital’s rule even when a function is not differentiable at the point of limit but is differentiable elsewhere in the interval. An example for that may be $\lim_{x\to0} x\ln|x|$ for which with little rearrangement we can apply the rule even when the functions $\ln|x|$ and $1/x$ are not differentiable at $x=0$.

  2. Secondly, the limit should be of the form $0/0$ or $\infty/\infty$. Well this is perhaps a well known condition for L’Hopital’s rule and can be explained from the proof of L’Hopital’s rule. An example for that could be $\lim_{x\to0}[x]/x$ where $[x]$ represents the greatest integer or floor function.

  3. The third restriction, i.e., $g'(x) \ne 0$, is necessary for the general proof of L’Hopital’s rule and is demonstrated beautifully by the counterexample provided by the Austrian mathematician Otto Stolz.

  4. Finally, the condition that $\lim_{x\to a}f'(x)/g'(x)$ must exist as illustrated by the counterexample $\lim_{x\to\infty} (x+\sin x)/x$.

So, basically the aforementioned four conditions determine if L’Hopital’s rule can be applied.

Now, there are many more examples where L’Hopital’s rule may not be the efficient way of solving the problem. There are countless examples where standard limits or series expansion provides an instant answer while L’Hopital’s rule becomes cumbersome and rather useless. Also, there have been cases where L’Hopital’s rule just gets into an infinite loop and won’t lead to an answer. https://youtu.be/66ixA-L3kVI

Toby Bartels
  • 2,131
  • 18
  • 34
Mathsmerizing
  • 1,903
  • 3
  • 17
1

(Edit: I focus on the case where $\underset{a}\lim f=\underset{a}\lim g=0$ (ie $f(a)=f(g)=0$ by extending $f$ and $g$ by continuity). But the other cases can be reduced to this case by considering $1/f$ and $1/g$)

It is worthwhile to remark that one of the only cases where the subtlety of L'Hopital's rule is actually needed is when $f$ or $g$ has a vertical tangent at $a$ (ie $f'$ or $g'\underset{a}{\rightarrow}±\infty$), which is rare. For instance : $\lim_{x\rightarrow1^-}\frac{\arccos(x)}{\sqrt{1-x}}$

In (almost) all other cases, when $f'(a)$ and $g'(a)$ exist and $g'(a)≠0$, the result is a mere consequence of the very definition of differentiability (namely, $f$ is differentiable at $a$$f(x)-f(a)\underset{x\rightarrow a}\sim (x-a)f'(a)$ ), since

$\frac{f(x)}{g(x)}=\frac{f(x)-\overbrace{f(a)}^{=0}}{g(x)-\underbrace{g(a)}_{=0}}\underset{x\rightarrow a}\sim\frac{(x-a)f'(a)}{(x-a)g'(a)}\underset{x\rightarrow a}\sim\frac{f'(a)}{g'(a)}$

(Edit: Even when $g'(a)=0$, it seems to me that L'Hopital's rule is often superfluous, because when $f$ and $g$ are differentiable enough times at $a$, one can apply Taylor's theorem to them, which is often faster than L'Hopital's rule which may require repeated derivatives.)

The beauty and subtlety of L'Hopital's rule stem from the fact that it doesn't require $f$ nor $g$ to be differentiable at $a$. Using this subtle and complex theorem in cases that are easily solvable with stronger assumptions and simpler properties is pedagogically absurd.

Edit: So, from my point of view, this is what explains the bad image of L'Hopital's rule: magical and often unproven so of little pedagogical value, and moreover quite useless compared to more efficient and easier to prove methods, except in some rare cases.

(It seems to me indeed that L'Hopital's theorem is especially popular in the U.S. curriculum compared to the rest of the world. For example, in France, it is totally absent from the curriculum, and is only marginally treated as an exercise. I find this much more consistent.)

Mr Jackie
  • 101
  • 6
  • If anyone thinks this is a low quality post, can they explain why and correct me? I have offered what seems to me to be a fairly coherent explanation of why L'Hopital's rule may be frowned upon by some, and why it may be pedagogically smart to avoid this magical and often undemonstrated rule in favor of other more accessible and understandable theorems, which is possible in most cases. – Mr Jackie Nov 01 '21 at 10:57
  • Your answer will have been reviewed by people specifically looking for new answers by new users to old posts, so they may have been predisposed to thinking that your answer is repetitive. Compared to the other new answer added yesterday, which went against the grain of the other answers, perhaps it is. But I disagree, since your answer is the only one to note that the subtlety and power of L'Hôpital's Rule is when the derivatives are ill-behaved. – Toby Bartels Nov 01 '21 at 12:31
1

I will go against the grain and say that no, there is no reason not to use l'Hopital's rule, in cases where the conditions for its use hold.

Claims of its "overuse," and frequency with which students are instructed "not to use l'Hopital's rule," puzzle me. (Especially when the intended solution is to Taylor expand the numerator and denominator---which is basically what l'Hopital amounts to!). One of the great beauties of mathematics is that a theorem, once proven, is ours to keep forever. If l'Hopital's rule is an extremely flexible mathematical tool that can solve a wide variety of problems, all the more reason to teach it and encourage students to become familiar with it.

Pedagogically, I understand the need for students to practice using other tools for evaluating limits, like the squeeze theorem etc. So... include problems in the homework that specifically require those tools (and where l'Hopital's fails)?

user7530
  • 45,846
  • 11
  • 84
  • 142