Evaluate $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$$ Without L'Hopital's Rule

$$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$

But I can not find a way to deal with $\sin x$ that does not result with a limit of the type $"\frac{0}{0}"$

  • $\begingroup$ Let $f(x)=\frac{x}{\sin x}$, then the limit equals $f'(0)$, but I don't think it helps... $\endgroup$ – Tony Ma May 14 '18 at 10:29
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    $\begingroup$ @TonyMa I can't see how the expression in the OP equals $\;f'(0)\;$ of your function... $\endgroup$ – DonAntonio May 14 '18 at 10:33
  • $\begingroup$ @DonAntonio Just devide both denominator and numerator by $\sin x$, hope I'm not wrong. (Maybe I forget to say that $f(0):=0$) $\endgroup$ – Tony Ma May 14 '18 at 10:35
  • $\begingroup$ @TonyMa Indeed you're right....and indeed you're even more right: that doesn't seem to help a lot. $\endgroup$ – DonAntonio May 14 '18 at 10:37
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    $\begingroup$ Very similar: math.stackexchange.com/questions/134051/… $\endgroup$ – Hans Lundmark May 14 '18 at 11:39

In 1800s trig texts it was common to prove without the use of calculus methods that for $0 < x < \frac{\pi}{2}$ we have

$$0 \; < \; x - \frac{1}{6}x^3 \; < \; \sin x \; < \; x $$

For a discussion of this, including 10 references (all of which can now be found freely available on the internet), see this 22 April 2008 ap-calculus post archived at Math Forum.

It follows that for $-\frac{\pi}{2} < x < 0$ we have

$$x \; < \; \sin x \; < \; x - \frac{1}{6}x^3 \; < \; 0 $$

Therefore, for all $x$ such that $-\frac{\pi}{2} < x < \frac{\pi}{2}$ we have

$$ | x - \sin x| \;\; < \;\; \left|x \; - \; \left(x - \frac{1}{6}x^3\right)\right| \;\; = \;\; \frac{1}{6}|x|^3 $$


$$ |\sin x| \; < \; |x| $$

Assuming these results, it follows that for all $x$ such that $\frac{\pi}{2} < x < \frac{\pi}{2}$ we have

$$ \left|\frac{x - \sin x}{x \sin x}\right| \;\; = \;\; \frac{|x - \sin x|}{|x \sin x|} \;\; < \;\; \frac{\frac{1}{6}|x|^3}{|x| \cdot |\sin x |} \;\; = \;\; \frac{|x|^2}{6|\sin x|} $$

$$ = \;\; \frac{1}{6} \cdot \left|\frac{x}{\sin x}\right| \cdot |x| \;\; \longrightarrow \;\; \frac{1}{6} \cdot 1 \cdot 0 \;\; = \;\; 0 $$

Incidentally, it is much easier to prove that for $\;0 < x < \frac{\pi}{2}\;$ we have $\;0 < x - \frac{1}{4}x^3 < \sin x < x,\;$ and this weaker result was also often proved without the use of calculus methods in older books. Note that we can still find the given limit using this weaker result --- just follow what I did above, changing all the $6$'s to $4$'s.

(ADDED A FEW HOURS LATER) Earlier I wrote, in regards to the references I gave in my 22 April 2008 post, that “all of which can now be found freely available on the internet”. I decided to google for them, along with some other papers I have copies of. For the most part I only considered papers, since I’d probably never finish trying to give all (or even most) of the 1800s textbooks that have a discussion of this inequality. For the few textbooks that I did include below, I used the earliest edition, since the entries are listed in chronological order. The links to the papers are to google-books versions and will take you to the first page of the paper. If there is a problem with any of them (such as needing a google account), then google the title for other locations where the paper can be found. For example, all of the French journals are at http://www.numdam.org. To make this more bibliographically useful, I tried to find full names and birth/death years for all authors.

[1] Alexandre Joseph Hidulphe Vincent (1797-1868), Note sur la construction des tables de sinus naturels, Nouvelles Annales de Mathématiques (1) 1 (1842), 272-277.

[2] François Joseph Eugène Lionnet (1805-1884), Sur une limite de l’erreur que l’on commet en remplaçant un arc par son sinus, Nouvelles Annales de Mathématiques (1) 2 (1843), 216-222.

[3] Auguste Deladéréere (??-??), Sur l’erreur commise en prenant un arc pour son sinus, Nouvelles Annales de Mathématiques (1) 2 (1843), 494-496.

[4] Olry Terquem (1782-1862), Théorème sur la difference entre l’arc et son sinus, Nouvelles Annales de Mathématiques (1) 3 (1844), 49-51.

[5] François Joseph Eugène Lionnet (1805-1884), Ueber eine für den elementar-unterricht in der trigonometrie vorzüglich geeignete methode zur erläuterung der berechnung der tafeln der sinus und cosinus, Archiv der Mathematik und Physik (1) 6 (1845), 205-213.

This is a “freely edited” (by the publisher) version of Lionnet’s 1843 paper above. Incidentally, the 3rd inequality on p. 209 should be $< \,.$

[6] Isaac Todhunter (1820-1884), Plane Trigonometry for the Use of Colleges and Schools, Macmillan and Company, 1859, vi + 271 pages.

See Article 120 on p. 83 for $\sin \theta > \theta - \frac{{\theta}^3}{4}.$ See Article 130 on pp. 88-89 for $\sin \theta > \theta - \frac{{\theta}^3}{6}.$

[7] Robert Rawson (1814-1906), Proof of the trigonometrical formula $\sin \theta > \theta - \frac{1}{6}{\theta}^3$, Messenger of Mathematics (old series) 3 (1866), 101-104.

[8] Joseph Joffroy (??-??), Démonstration de la formule $a - \sin a < \frac{a^3}{4}$, Nouvelles Annales de Mathématiques (2) 8 (1869), 42-43.

[9] François Joseph Eugène Lionnet (1805-1884), Note sur les questions 1045 et 1026, Nouvelles Annales de Mathématiques (2) 11 (1872), 78-81.

See Remarque on p. 81, which states that $a - \sin a < \frac{a^3}{6}$ and $1 - \cos a < \frac{a^2}{2}$ follow from the (geometrically proved) theorems 1 and 2.

[10] Joseph Joffroy (??-??), Démonstration géométrique de l’inégalité $a - \sin a < \frac{a^3}{4}$, Nouvelles Annales de Mathématiques (2) 14 (1875), 171-172.

[11] François Joseph Eugène Lionnet (1805-1884), Sur une limite de l’erreur, Journal de Mathématiques Élémentaires (1) 3 (1879), 193-197.

See IV. Théorème on pp. 196-197.

[12] [author not given], Démontrer élémentairement que l’on a $x - \sin x < \frac{x^3}{6}$, Journal de Mathématiques Élémentaires et Spéciales (1) 5 (1881), 156-157.

This is a published solution to an examination problem.

[13] Joseph Edwards (1854-1931), Differential Calculus with Applications and Numerous Examples. An Elementary Treatise, Macmillan and Company, 1886, xvi + 439 pages.

See Article 34 on pp. 22-23.

[14] Louis Desmons (1850-1921), Démonstration élémentaire de l'inégalité $x - \sin x < \frac{x^3}{6}$, Journal de Mathématiques Élémentaires (3) 3 (1889), 145-146.

[15] Louis Desmons (1850-1921) and Émile Gelin (1850-1921), Démonstration élémentaire de l'inégalité $x - \sin x < \frac{1}{6}x^3$, Mathesis Recueil Mathématique (1) 10 (1890), 58-60.

[16] Pierre Maximilien Évariste Bernès (1831-??), Démonstration de l'inégalité $\sin x > x - \frac{1}{6}x^3$, Mathesis Recueil Mathématique (1) 10 (1890), 112-113.

[17] J. Smeets (??-??), Sur l’inégalité $x - \sin x < \frac{1}{4}x^3$, Mathesis Recueil Mathématique (1) 10 (1890), 157-158.

The REMARQUE at the end, on p. 158, is by Joseph Jean Baptiste Neuberg (1840-1926).

[18] Paul Alexandre Pierre Delens (1856-??), Théorème de trigonométrie, Mathesis Recueil Mathématique (2) 4 (1894), 68-69.

[19] A. Absolonne (??-??), [Solution to Question 839], Mathesis Recueil Mathématique (2) 4 (1894), 73-76.

This answers a question proposed by Francesco Giudice (1855-1936) by giving a proof by elementary considerations that $\tan x - x < \frac{1}{3}{\tan}^3x.$ The NOTE on pp. 75-76 is by Joseph Jean Baptiste Neuberg (1840-1926).

[20] Maurice Fouché (1855-1929), Démonstration de l’inégalité $x - \sin x < \frac{1}{4}x^3$, Mathesis Recueil Mathématique (2) 5 (1895), 117.

[21] Josef Krug (??-??), [Untitled note], Archiv der Mathematik und Physik (3) 12 (1907), 92.

[22] Charles Davison (1858-1940), Subjects for Mathematical Essays, Macmillan and Company, 1915, x + 160 pages.

See Section 78. The Inequality Theorem $\sin \theta > \theta - \frac{1}{6}{\theta}^3$ on pp. 84-85.

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    $\begingroup$ It might be easier to say $0\lt x-{1\over6}x^2\lt\sin x\lt x$ implies $$0\lt{1\over\sin x}-{1\over x}\lt{1\over x-{1\over6}x^3}-{1\over x}={x\over6-x^2}\to0$$ for $x\gt0$, and the function is odd. $\endgroup$ – Barry Cipra May 14 '18 at 12:47
  • $\begingroup$ @Barry Cipra: Interestingly, I originally got $\frac{x}{6 - x^2},$ but then realized I applied one of the inequalities backwards (the inequality above that I never used, because I forgot to delete it, namely $|\sin x| < |x|).$ However, your method of dealing with the reciprocals seems to make it work. $\endgroup$ – Dave L. Renfro May 14 '18 at 12:54

A rigorous treatment of trig functions would necessitate the use of power series, when this becomes trivial (with all the tools that come along with the theory of converging power series). Let me proffer the following argument relying on looking up things from a picture.

The function is odd, so it suffices to handle the case $x>0$. Let's look at the part of the unit circle: $O=(0,0)$, $A=(1,0)$, $P=(\cos x, \sin x)$.

enter image description here

The triangle $\Delta OAP$ has area $\dfrac12\sin x$. That triangle is contained in the sector $\angle OAP$ with area $\dfrac12 x$ ($x$ is the length of the arc $AP$). Therefore $0<\sin x<x$. Presumably all are familiar with this estimate. I will use it below without mentioning it explicitly.

Furthermore, the thin circular segment in between the arc $AP$ and the line segment $AP$ has area $(x-\sin x)/2$.

The line segment $AP$ has length $2\sin(x/2)$, and the maximum height of the circular segment is obviously (along the red line in the image) $h=1-\cos(x/2)$. Therefore the area of the segment is bounded from above by $2h\sin(x/2)$. This implies the inequalities $$ 0<\frac{x-\sin x}{x\sin x}<\frac{4\sin (x/2)(1-\cos(x/2))}{x\sin x}. $$ Let's work on that upper bound. In the denominator we can write the sine as $$ \sin x=2\sin(x/2)\cos(x/2). $$ Multiplying both the numerator and the denominator by $1+\cos(x/2)$ then gives the upper bound $$ \begin{aligned} \frac{4\sin (x/2)(1-\cos(x/2))}{x\sin x}&=\frac{4\sin^3(x/2)}{2x\sin(x/2)\cos(x/2)(1+\cos(x/2))}\\ &<\frac{x}{2\cos(x/2)(1+\cos(x/2))}\\ &<x \end{aligned} $$ for all $x\in(0,\pi/3)$ (when $2\cos(x/2)$ exceeds $1$). The squeeze theorem then implies that the limit as $x\to0+$ is equal to zero.


Let $0<x<\pi/2.$ Consider the sector within the unit circle with vertices $(0,0),(1,0),(\cos x,\sin x);$ its area is $x/2.$ That area is smaller than the area of the right triangle with vertices $(0,0),(1,0),(1,\tan x),$ whose area is $(\tan x)/2.$ (Good to draw a picture.) This gives the inequality $x < \tan x.$ Thus

$$0< x-\sin x < \tan x - \sin x = \tan x (1-\cos x)$$ $$ < \tan x (1-\cos^2 x) = \tan x \cdot \sin^2 x.$$


$$0< \frac{x-\sin x}{x^2} < \tan x \frac{\sin^2 x}{x^2}.$$

As $x\to 0^+,$ $\tan x \to 0$ and $(\sin^2 x)/x^2 \to 1.$ Thus our limit, from the right, is $0.$ Because our function is odd, the limit is also $0$ from the left, and we're done.

  • $\begingroup$ Good job! Simpler than mine. $\endgroup$ – Jyrki Lahtonen May 15 '18 at 17:28
  • $\begingroup$ @JyrkiLahtonen Thank you. $\endgroup$ – zhw. May 15 '18 at 17:36

Using a taylor expansion for $\sin(x)$, you can find the limit.

Hint: $\sin(x) = x − \frac{x^3}{3!} + \frac{x^5}{5!} −O(x^7)$

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    $\begingroup$ Taylor expansions use derivatives...so then why not apply directly L'Hospital? $\endgroup$ – DonAntonio May 14 '18 at 10:27
  • $\begingroup$ Downvoting is too much for any answer (even an incorrect one, and this one is correct!). Perhaps giving some slack and wait would be better...? $\endgroup$ – DonAntonio May 14 '18 at 10:30
  • $\begingroup$ @Don I don't know why on this site sometimes people prefer LHospital ans other times taylor. This answer is, as you said perfectly alright. I guess you want to say that its easier to apply LH here? $\endgroup$ – jonsno May 14 '18 at 10:47
  • $\begingroup$ Using l'Hôpital is easy, but as an exercise it is perfectly normal to ask students not to use the most straightforward way but instead use taylor expansions. I think this is an alternative way so students learn to seek multiple ways to solve a problem. $\endgroup$ – Fibonacci May 14 '18 at 10:49
  • $\begingroup$ In this case L’Hopital need to be used twice so I am trying to find another way $\endgroup$ – gbox May 14 '18 at 10:53

Given your steps, without derivates and Taylor, $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$

an knowing that $$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\frac16$$

you have:

$$\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}=\lim_{x\to 0}\frac{x^2}{x^2}\left(\frac{1-\frac{\sin x}{x}}{\sin x}\right)=\lim_{x\to 0}\left[\underbrace{\frac{1-\tfrac{\sin x}{x}}{x^2}}_{\to \frac 16}\left(\underbrace{\frac{x}{\sin x}}_{\to 1}\right)\right]\, x=0$$


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