I'm looking for cases like $$\lim_{x \to 0} \frac {1-\cos(x)}{x^2}$$ that will not give you the answer the first time you use L'Hôpital's rule on them. For example in this case it will result in a number $\frac{1}{2}$ the second time you use L'Hôpital's rule. I want examples of limits like $\lim_{x \to c} \frac {f(x)}{g(x)}$ so that you have to use L'Hôpital's rule $5$ times, $18$ times, or say $n$ times on them to get an answer. Another question is about the case in which you use L'Hôpital's rule as many times as you want but you always end with $\lim_{x \to 0} \frac {0}{0}$. Does this case exist?

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    $\begingroup$ The spirit of L'Hospital's Rule is not to apply it repeatedly. Rather it is applied in the hope that perhaps the limit of $f'(x) /g'(x) $ is easier to evaluate that evaluating the original limit. Multiple applications of L'Hospital's Rule are nothing but just using Taylor's theorem instead and it is much easier to apply Taylor directly. $\endgroup$ – Paramanand Singh Nov 10 '17 at 18:38
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    $\begingroup$ You could apply "L'Hopital" twice to find the limit of $\frac{1- cos(x)}{x^2}$ but you don't have to. Most people would apply "L'Hopital" once to get $\frac{sin(x)}{2x}= \frac{1}{2}\frac{sin(x)}{x}$ and you should have learned that $\frac{sin(x)}{x}$ has limit 1 as x goes to 0. $\endgroup$ – user247327 Nov 10 '17 at 19:49
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    $\begingroup$ Does $\displaystyle\lim_{x \to \infty}x^ne^{-x}$ count as an example? $\endgroup$ – JimmyK4542 Nov 11 '17 at 3:27
  • $\begingroup$ How about $\lim_(n \to \infty) \dfrac{\sin(\tan x)}{x^6}$? $\endgroup$ – ExtremeRaider May 17 at 15:12

Sure. Do you want $18$ times? Then consider the limit$$\lim_{x\to0}\frac{x^{18}}{x^{18}}$$or the non-trivial example$$\lim_{x\to0}\frac{\sin(x^{18})}{1-\cos(x^9)}.$$For the case in which you always get $\frac00$, consider the function$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}e^{-1/x^2}&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}\end{array}$$and the limit$$\lim_{x\to0}\frac{f(x)}{f(x)}$$or the non-trivial example$$\lim_{x\to0}\frac{f(x)}{f(x^2)}.$$

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    $\begingroup$ @MichaelSeifert Do you think that$$\lim_{x\to0}\frac{\sin(x^{18})}{1-\cos(x^9)}$$is in the spirit of thisquestion? $\endgroup$ – José Carlos Santos Nov 10 '17 at 17:50
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    $\begingroup$ @JoséCarlosSantos I can't speak to Michael's or the OP's tastes, but to me, that's much more interesting because it can't be trivially resolved through algebra. $\endgroup$ – Aaron Montgomery Nov 10 '17 at 21:25
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    $\begingroup$ @AaronMontgomery: Why not? As $t \to 0$ you have $\sin(t)/t \to 1$ and $(1-\cos(t))/t^2 \to 1/2$. Using these with the appropriate $t$ yields an 'algebraic' solution to your limit. $\endgroup$ – user21820 Nov 11 '17 at 14:15
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    $\begingroup$ @user21820 I wouldn't call the limit of $sin(t)/t$ a trivial algebraic limit; I'd call it a nontrivial geometric limit. $\endgroup$ – Aaron Montgomery Nov 11 '17 at 14:18
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    $\begingroup$ @user21820 I'm thinking of this problem in elementary terms, i.e. how it would be approached in early calculus. The $x^n/x^n$ limit would be trivially handled by a student in the second or third week of calculus; the $\sin(x) / x$ limit would not. $\endgroup$ – Aaron Montgomery Nov 11 '17 at 16:35

A couple of rather famous limits that each require 7 applications of L’Hôpital’s rule (unless evaluated by another method) are

$$ \lim_{x \rightarrow 0} \,\frac{\tan{(\sin x)} \; - \; \sin{(\tan x)}}{x^7} \;\;\; \text{and} \;\;\; \lim_{x \rightarrow 0} \, \frac{\tan{(\sin x)} \; - \; \sin{(\tan x)}}{\arctan{(\arcsin x)} \; - \; \arcsin{(\arctan x)}} \;\; $$

These two limits are discussed in the chronologically listed references below, with [11] being a generalization of the tan/sin and arctan/arcsin version. (Both [10] and [11] were brought to my attention by user21820.) Another limit that requires 7 applications of L’Hôpital’s rule is the following, which I mentioned (in an incorrect way, however) at the end of [6]:

$$ \lim_{x \rightarrow 0} \,\frac{\tan x \; – \; 24\tan \frac{x}{2} \; - 4\sin x \; + \; 15x}{x^7} $$

[1] sci.math, 13 February 2000

[2] sci.math, 16 April 2000

[3] sci.math, 11 July 2000

[4] sci.math, 13 August 2001

[5] sci.math, 12 February 2005

[6] sci.math, 27 December 2007

[7] sci.math, 7 October 2008

[8] A question regarding a claim of V. I. Arnold, mathoverflow, 8 April 2010.

[9] How find this limit $\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$, Mathematics Stack Exchange, 2 November 2013.

[10] Limit of $\dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$ as $x \rightarrow 0$, Mathematics Stack Exchange, 26 May 2014.

[11] $\lim_{x \to 0} \dfrac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)} = 1$ for any $f,g \in C^1$ that are tangent to $\text{id}$ at $0$ with some simple condition, Mathematics Stack Exchange, 26 May 2014.

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    $\begingroup$ Your limit at the end just needs one application of L'Hospital's Rule. The denominator changes to $4\cos x-15$ and numerator changes to $\sec^{2} x-12\sec^{2}(x/2)$. Both of these expressions tend to $-11$ so that the limit is $1$. Am I missing something or have you made a typo? $\endgroup$ – Paramanand Singh Nov 10 '17 at 18:32
  • $\begingroup$ @Paramanand Singh: You're correct. Using a computer algebra system, I just now found their power series expansions about $x=0,$ and each begins with $-11x,$ so one application of L'Hopital's rule will do it. However, the power series agree until the $x^7$ term, so their difference equals $O(x^7)$ (means: approaches $0$ at least as fast as a constant times $x^7).$ Apparently I saw this agreement in their power series expansion in some book or paper (might have been Hardy's A Course of Pure Mathematics) and I incorrectly interpreted the result when I made my comment at the end of [6]. $\endgroup$ – Dave L. Renfro Nov 10 '17 at 22:06
  • $\begingroup$ Ok so the more interesting / hard limit is $$\lim_{x\to 0}\frac{\tan x-24\tan(x/2)-4\sin x+15x}{x^{7}}$$ $\endgroup$ – Paramanand Singh Nov 10 '17 at 22:10
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    $\begingroup$ @Paramanand Singh: The limit loses quite a bit of its appeal to me now, because it's relatively easy (I think) to find something like this by trial and error with power series expansions. That is, arrange for two functions (each made up of several algebraic and/or transcendental terms) to have the first few terms of their expansions agree, then just divide their difference by the lowest power of $x$ where their expansions differ. Much more interesting, I think, is when you have something like what I had before (if it had been valid), where it's not obvious how one might have discovered it. $\endgroup$ – Dave L. Renfro Nov 10 '17 at 22:19
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    $\begingroup$ Would you like to add this and this related question to your laundry list? In particular, the latter gives somewhat a generalized form that is not just accidental or ad-hoc. =) $\endgroup$ – user21820 Nov 11 '17 at 8:33

To me, the simplest (nontrivial) way to do this is to exploit functions' representations as power series. For instance, begin with:

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots = \sum_{i=0}^{\infty} \frac{x^i}{i!}.$$

To cook up an interesting L'Hospital problem, subtract off the first few terms of this series expansion from $e^x$ and divide by an appropriate term. All the following are classical Calculus I examples which are inspired by the above series expansion: \begin{align*}\lim_{x \to 0} \frac{e^x - 1}{x} &\qquad \text{(requires 1 use of L'H)} \\ \lim_{x \to 0} \frac{e^x - 1- x}{x^2} &\qquad \text{(requires 2 uses of L'H)} \\ \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3} &\qquad \text{(requires 3 uses of L'H)} \end{align*} and so forth. You can pick any function you like in place of $e^x$, of course, so long as it has enough derivatives to play with.

You can also use this approach to cook up slightly more interesting examples. For instance, we could subtract off the appropriate terms from $e^x$ and $\cos(x)$ to get their series expansions to be $Cx^2 + [\text{lower order terms}]$. Specifically, $$\lim_{x \to 0} \frac{e^x - 1 - x}{\cos(x)- 1 }$$ has a nonzero limit and requires two uses of L'Hospital's rule. If we wanted four, we could have subtracted out the $x^2$ and $x^3$ terms from the $e^x$ expansion and the $x^2$ term from the $\cos(x)$ expansion.

What I like about this approach:

  1. The examples are nontrivial, in the sense that no elementary algebraic techniques will save you from having to use L'Hospital's rule.
  2. You can immediately tell how many uses of L'Hospital's rule will be required.
  3. I think it conveys something important both about Taylor series representations of functions and about how L'Hospital's rule works.
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    $\begingroup$ You could argue that this is, to some extent, the definition of a Taylor polynomial. $\endgroup$ – AccidentalFourierTransform Nov 10 '17 at 22:37
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    $\begingroup$ This is a very useful answer in that it shows why one would have to use the rule multiple times. Each application gets rid of one power of the variable in both the numerator and denominator. If you cancel lots of terms, you have high powers of the variable and have to use the rule lots of times. $\endgroup$ – Ross Millikan Nov 11 '17 at 3:52
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    $\begingroup$ This is a wonderful answer. +1 for imparting (potential) insight about Taylor series and about L'Hospital at the same time! $\endgroup$ – G Tony Jacobs Nov 11 '17 at 15:29

For the "$\frac{\infty}{\infty}$" case, if you only use the L'Hôpital's rule and don't change your fraction between its successive applications, then one of simple nontrivial never-ending examples from many textbooks is $$\lim_{x\to0+}\frac{\ln x}{\cot x}$$.


You can construct simple (boring) examples quite easily using polynomials. A very trivial example is $\frac{x^n}{x^n}$. For the never works case, replace $x^n$ with the interesting function: $e^{\frac{-1}{x^2}}$.


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