Often times, users on this page will ask about how to evaluate this or that limit without using L'Hospital's rule.

Lately, these questions are tagged (more and more, although mostly by editors, but hey, that's a start) with the tag to differentiate them from "regular" limit questions. And the questions get their share of answers (or close votes, depending on the quality).

However, the problem is that a vast majority of answers, at least by my experience, tend to use Taylor's expansion to quickly evaluate the limit. The example can be seen here:

Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule

I believe that answering a question where L'Hospital is not to be used by using Taylor's formula is cheating and not useful to the OP, and I therefore downvote all answers using Taylor. It is like being forbidden to use pistols in a duel, and then bringing a cannon to the match. Taylor's expansion, if it works, does the exact same thing as L'Hospital's rule, i.e. it replaces calculating $f/g$ with $f'/g'$.

What's worse, usually (not in the example I provide), authors of answes are usually pretty angry at the downvotes, claiming "this works, why would you downvote?" and not accepting my explanation at all.


My question is this:

Do you agree that answers using Taylor's expansion for limits without L'Hospital are not useful (and it is therefore correct to downvote them)?

  • 1
    $\begingroup$ @Behaviour You're right, I changed my question, but the point remains. however, I don't agree with downvoting the questions. Often times, they are posted by people who worked on the question and just need some help, and saying "use Taylor" is not helping. $\endgroup$
    – 5xum
    Commented Dec 22, 2014 at 14:46
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    $\begingroup$ The general issue of the "without LH" is tricky. I am in favor of deciding on a case-by-case basis. (Also, I just saw something so much worse here, so I am soft on this right now.) [My upvote is for the idea to discussing this, not to agree with the proposal.] $\endgroup$
    – quid Mod
    Commented Dec 22, 2014 at 14:47
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    $\begingroup$ I hate these kinds of questions because they're ill-defined. Rarely is it explicitly stated what you are or are not allowed to assume has already been proven about limits, and never is it clear why the "best" answer isn't just to first prove l'Hopital with those tools. $\endgroup$
    – user7530
    Commented Dec 22, 2014 at 15:44
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    $\begingroup$ Such downvoting is nonconstructive. Instead, I'd imply relay this viewpoint to the OP and ask the OP to clarify if they are interested in solutions using (Taylor) series expansions. $\endgroup$ Commented Dec 22, 2014 at 15:52
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    $\begingroup$ In my opinion, the Maclaurin expansion approach is very different in spirit from the L'Hospital's Rule approach, since it confronts directly the behaviour of the component functions near $0$. As to immediate usefulness to the OP, she can judge among the answers. Some will undoubtedly use "tricks" that reduce the problem to a familiar limit. $\endgroup$ Commented Dec 22, 2014 at 15:57
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    $\begingroup$ It does not happen often I essentially agree with Bill Dubuque and André Nicolas. But, there we have it. :-) $\endgroup$
    – quid Mod
    Commented Dec 22, 2014 at 16:08
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    $\begingroup$ I'm inclined to downvote such answers as well, but for different reasons. You can tell something about where the OP is in their mathematics education when they ask a question like this. At such a stage, a series solution to the limit is almost certainly not appropriate. Helpful for intuition, maybe. But for the formal answer, no. $\endgroup$
    – Simon S
    Commented Dec 22, 2014 at 16:51
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    $\begingroup$ No, I certainly do not agree. Frankly, I consider such downvotes sheer vandalism when the answer is correct and reasonably clearly written. Such an answer may or may not be useful to the person asking that particular question, but it may well be useful to someone reading the page. If you question its appropriateness, you can leave a comment to that effect — or, better yet, provide a better answer. $\endgroup$ Commented Dec 25, 2014 at 22:40
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    $\begingroup$ @BrianM.Scott I can assure you you are wrong in at least one thing: the downvotes, at least from my side, are not "vandalism". They may be misguided (by your oppinion), but they are given because I believe (or, at least, believed) that giving them is the right thing to do. $\endgroup$
    – 5xum
    Commented Dec 26, 2014 at 7:27
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    $\begingroup$ @5xum: Your belief that you're doing the right thing is irrelevant: vandalism with the best of intentions is still vandalism. $\endgroup$ Commented Dec 26, 2014 at 9:20
  • $\begingroup$ I would have to agree with Brian M. Scott. How is Taylor's Expansion the same as L'hopital's rule? You say it is similar to $f'/g'$ consider $f(x) = e^x$ how in the world does $\displaystyle 1 + \sum_{n=1}^{\infty} x^n/n!$ look like $e^x$? That is like saying an iPhone looks like a Samsung.... $\endgroup$
    – Amad27
    Commented Dec 29, 2014 at 15:13
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    $\begingroup$ Some books (Rudin) use Taylor series in fact as the definition of $e^x$. (Similar point was given in user48672's answer.) $\endgroup$ Commented Dec 30, 2014 at 14:37
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    $\begingroup$ @Amad27, though the discussion is starting to get off-topic, here's the general idea. If $f,g$ are analytic at the origin and $f(0) = g(0) = 0$ with $g'(0) \neq 0$, then $$\frac{f(x)}{g(x)} = \frac{f'(0)}{g'(0)} + O(x)$$ as $x \to 0$, which is L'Hopital's rule in this particular case. $\endgroup$ Commented Dec 30, 2014 at 18:07

3 Answers 3


Downvotes might be appropriate in some cases (where Taylor is extreme overkill), but in general I disagree. For example, in the calculus courses at my university, we avoid l'Hospital for various pedagogical reasons, most of which have been discussed in this question; see also comments here. (And I might add my pet peeve that this theorem is almost always invoked without bothering to verify all the necessary hypotheses, in particular that the derivative of the denominator is nonzero in a punctured neighbourhood of the point in question.) Instead, we advocate using standard limits like $\lim_{x \to 0} (\sin x)/x=1$, or Taylor series in more complicated situations. For a student taking such a course, an answer in terms of Taylor series would be a perfectly valid response to a "do it without l'Hospital" question.

  • $\begingroup$ But don't you think that a "do it without l'Hospital" question usually appears to students who don't yet know Taylor? In that case, an answer using Taylor is useless to them... $\endgroup$
    – 5xum
    Commented Dec 23, 2014 at 6:24
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    $\begingroup$ I guess it depends on the traditions of the educational system where you happen to live. In the situation that I'm used to (in Sweden), it's not so, but it seems to me that in North America (for example) there is a lot more emphasis on l'Hospital, and in that case you're probably right. $\endgroup$ Commented Dec 23, 2014 at 10:12
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    $\begingroup$ I think that it is relevant to mention that you do not need to know Taylor's theorem (or any kind of differentiation) to be able to use Taylor series representations for some functions. For example in the analysis course at my university, the exponential function, sine and cosine are defined by their MacLaurin series long before differentiation is introduced. Hence, such an answer could be relevant for the respective OP. $\endgroup$
    – PhoemueX
    Commented Dec 27, 2014 at 17:17

Personnally I find it annoying when the asker requests a solution not using Taylor expansions, and 20 seconds later someone rushes to post a solution using them. It feels like the answerer cares not at all about mathematics and pedagogy, but only about reputation points, an attitude which is in my opinion not helpful and quite absurd. In those situations I feel entitled to express my discontent by downvoting them.

But if the asker doesn't precise anything, I don't think they should be downvoted, since it is possible that the answerer had honest intentions, and it would be unjust to punish them. Even if I totally agree with your metaphor comparing L'Hopital to pistols and Taylor to cannons.

About the parallel discussion of the definitions of trigonometric and exponential functions: in high school here in France, $\exp$ is defined as the inverse function $\ln$, which in turn is uniquely defined by $\ln'(x)=\frac{1}{x}$ and $\ln(1)=0$. Trigonometric functions are defined geometrically, and if you want to give those definitions a more formal flavor, you can proceed similarily: $\sin$ is the inverse function of $\arcsin$ which is defined uniquely on $[-1,1]$ by $\arcsin'(x)=\frac{1}{\sqrt{1-x^{2}}}$ and $\arcsin(0)=0$.

I find these definitions far more reasonable than using power series. Even though it is possible to define $\pi$ as half the period of the function $x\mapsto e^{ix}$, historically $\pi$ is the perimeter of a circle with width $1$, and I'm sure most of us think of $\pi$ like that. In the same way, $\ln$ was introduced before $\exp$, and $e$ was defined by Leibniz as the solution of $\ln(x)=1$, before Euler discovered $\exp$'s power series.

Just because definitions using power series are powerful and seem easy doesn't mean they are the most faithful to the nature of the mathematical objects involved. There is more to mathematics than pure formal proofs and definitions: there is understanding of the ideas behind those proofs and definitions. Each time someone unconcernedly posts a Taylor solution to a without l'Hopital question it seems to me that he cares only about the former and not about the latter. I hope that I am wrong most of the time.


I don' know why you would not use Taylor. One of the definitions of $\sin(x)$ is the sum of the odd powers in the expansion of $\exp (i x) $. If you don't agree please provide your definition of sin(x).

  • $\begingroup$ I can say immediately claim that the example is divergent to $\infty$ from the regular sinc(x) definition. $\endgroup$
    – user48672
    Commented Dec 22, 2014 at 23:41
  • $\begingroup$ $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$, $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, $\lim_{x\to0}\frac{\sin x}x=1$, and continuity, uniquely define $\sin$ and $\cos$, I believe. (Leaving out the last two equalities uniquely define $\sin$ and $\cos$ without units—that is, $\sin(x^\circ)$ and $\cos(x^\circ)$ would also work.) All of these properties can be proven geometrically. $\endgroup$ Commented Dec 23, 2014 at 3:15
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    $\begingroup$ This answer, sorry to say, is exactly the type of answer I usually downvote. When you first learned of $\sin$ and calculated it's limits, what definition did you use? I bet it wasn't the "sum of the odd powers in the expansion of $exp(ix)$"... Also, the example is not divergent. $\endgroup$
    – 5xum
    Commented Dec 23, 2014 at 7:07
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    $\begingroup$ @5xum: You are not correct. As I stated in the comment to the other answer, analysis courses at my university usually introduce the exponential function, sine and cosine directly by their power series, ignoring any further geometric intuition. I quite like that approach. Hence, answers using these series would be very helpful for the OP if he is taking such a course. $\endgroup$
    – PhoemueX
    Commented Dec 27, 2014 at 17:20
  • $\begingroup$ @columbus8myhw: Your first two equations would certainly also hold for $\sin \equiv \cos \equiv 0$. $\endgroup$
    – PhoemueX
    Commented Dec 27, 2014 at 17:22
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    $\begingroup$ @PhoemueX I later discovered that those conditions do not uniquely define sine and cosine, but adding any one of (a) $\sin^2(x)+\cos^2(x)=1$, (b) $\sin$ is odd and $\cos$ is even, or (c) $\lim_{x\to0}\frac{\cos(x)-1}x=0$ is enough to uniquely define them. $\endgroup$ Commented Dec 27, 2014 at 22:53

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