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How can one prove the statement $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.

This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.

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l'Hôpital's rule is easiest: $\lim\limits_{x\to0}\sin x = 0$ and $\lim\limits_{x\to0}x = 0$, so $\lim\limits_{x\to 0}\frac{\sin x}x = \lim\limits_{x\to 0}\frac{\cos x}1 = 1 $ – Joren Oct 23 '11 at 20:41
@Joren: I'm extremely curious how will you prove then that $\sin ' x = \cos x$ – Ilya Oct 24 '11 at 9:10
@FUZx44xl: sure, but to be fare you first prove that $\sin x\sim x$ with $x\to 0$. Geometrically – Ilya Oct 24 '11 at 15:42
It depends on how you define $\sin$! – user59671 May 9 '13 at 11:24
Recent changes in what? The definition of $\lim$, of $\sin$ or of $0$? – Asaf Karagila Dec 18 '13 at 20:03

17 Answers 17

up vote 274 down vote accepted

sinc and tanc at 0

The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$

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Do my homework! Note that $(1)$ says that for $0\le x\le\frac{\pi}{2}$ we have $0\le\sin(x)\le x$; therefore, $\lim\limits_{x\to0^+}\sin(x)=0$. Then $\cos(x)=1-2\sin^2(x/2)$ should finish the job. – robjohn Oct 23 '11 at 18:37
Thank you very much. I know that proverb, but I really wasn't able to find that out on my own. – FUZxxl Oct 23 '11 at 18:41
From your comment, I wasn't expecting that you could find it on your own, but "Read the question!" seemed a bit rough around the edges. – robjohn Oct 23 '11 at 19:04
Sorry for that. – FUZxxl Oct 23 '11 at 19:35
I think that your justification is slightly circular :). In an introductory calculus course, $(\sin x,\cos x)$ is probably defined to be the point we reach after traveling $x$ units counterclockwise along the unit circle from $(1,0)$. – Mike F Oct 23 '11 at 20:54

You should first prove that for $x > 0$ small that $\sin x < x < \tan x$. Then, dividing by $x$ you get $$ { \sin x \over x} < 1 $$ and rearranging $1 < {\tan x \over x} = {\sin x \over x \cos x }$ $$ \cos x < {\sin x \over x}. $$ Taking $x \rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $\sin(-x) = -x$ so that ${\sin(-x) \over -x} = {\sin x \over x}$.

As far as why the first inequality I said is true, you can do this completely from triangles but I don't know how to draw the pictures here.


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But how to prove that $\sin x<x<\tan x$? – FUZxxl Oct 23 '11 at 16:31
It is in the picture. The definition of radians makes the picture above true. Maybe that is worth mentioning: this limit explicitly depends on "$x$" being measured in radians. – tkr Oct 23 '11 at 16:33
Okay. I had a look at the link Yuval provided. That proof works. Anyway, thanks for the effort. – FUZxxl Oct 23 '11 at 16:34
This is a strange picture! Normally you want the $tan(\theta)$ side to be parallel to the $sin(\theta)$ side. – user641 Oct 23 '11 at 17:36
If you make $\tan(\theta)$ parallel then you need to make the points $S$ and $Q$ the same. For whatever reason, this is the proof I like the most because it relates the tangent line at the point on the circle to the value we call "tangent". To each his own... – tkr Oct 23 '11 at 18:22

Usually calculus textbooks do this using geometric arguments followed by squeezing.

Here's an Euler-esque way of looking at it---not a "proof" as that term is usually understood today, but still worth knowing about.

Let $\theta$ be the length of an arc along the circle of unit radius centered at $(0,0)$, from the point $(1,0)$ in a counterclockwise direction to some point $(\cos\theta,\sin\theta)$ on the circle. Then of course $\sin\theta$ is the height of the latter point above the $x$-axis. Now imagine what happens if $\theta$ is an infinitely small positive number. Then the arc is just an infinitely short vertical line, and the height of the endpoint above the $x$-axis is just the length of the arc. I.e. when $\theta$ is an infinitely small number, then $\sin\theta$ is the same as $\theta$. It follows that when $\theta$ is an infinitely small nonzero number, then $\dfrac{\sin\theta}{\theta}=1$.

That is how Euler viewed the matter. See his book on differential calculus.

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I disagree with the recent edit to my answer and I have reverted to the previous version. "Infinitesmal" means "infinitely small". $\qquad$ – Michael Hardy Jan 21 at 19:12
You read Euler's book ? Was it very difficult to read because of the notation and language of the time ? – user230452 Feb 27 at 4:03
@user230452 : Just some parts of it. I wouldn't say the differences in language and notation were the challenging part. $\qquad$ – Michael Hardy Feb 27 at 18:18

Look this link

Here is the picture copied from that blog:

Copy of the picture from the Fatos Matematicos blog

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+1, nice site - What is the length BC? – NoChance Oct 24 '11 at 5:34
+1 same idea as my proof, but much cleaner. – John Joy Nov 9 '14 at 15:07
What is the argument for showing that $\theta \cos \theta\le \sin \theta$? The picture isn't immediately convincing. – Dr. MV Apr 15 '15 at 5:31
@Dr. MV: if you rescale the circle in $\cos \theta$ units, then the original $\sin \theta$ is a scaled $\tan \theta$ which is always longer than its arc. – Mattsteel May 30 at 20:38

Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. When we take $\lim_{n\rightarrow\infty} \frac{n\sin(\frac{\pi}{n})}{1+\sin(\frac{\pi}{n})}$ we may notice that we have infinitely many circles surrounding the unit circle with infinitely small diameters that lastly perfectly approximate the length of the unit circle when having it there infinity times . Therefore when multiplying n by the radius under the limit to infinity we get π. Let's denote $\frac{\pi}{n}$ by x.


The proof is complete.

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this assumes the a priori knowledge of the existence of the limit $\lim_{x\to 0}\frac{\sin(x)}{x}$, how would you go about proving it before hand? (+1) by the way for the original and alternative proof – user153330 Feb 24 at 14:07

I am not sure if it counts as proof, but I have seen this done by a High Schooler. enter image description here

In the given picture above, $\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of polygon }$.

$\displaystyle \lim_{n\to \infty }2nR \sin\left( \frac{\pi}{n } \right ) = \lim_{n\to \infty } (\text{ perimeter of polygon }) = 2 \pi R \implies \lim_{n\to \infty}\frac{\sin\left( \frac{\pi}{n } \right )}{\left( \frac{\pi}{n } \right )} = 1$ and let $\frac{\pi}{n} = x$.

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This method is usually used to prove that the perimeter of a circle is $2\pi R$ using the fact $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$ – whatever Jul 20 '13 at 19:15
Santosh, how does one prove that the perimeter of the polygon converges to the $ 2\pi$ ? – Imago Feb 5 at 19:45

I claim that for $0<x<\pi/2$ that the following holds $$\sin x \lt x \lt \tan x$$ Figure 1
In the diagram, we let $OC=OA=1$. In other words, $Arc\:CA=x$ is an arc of a unit circle. The shortest distance from point $C$ to line $AO$ is line $CE=\sin x$ (because $CE\perp OA$). Another path from point $C$ to line $OA$ is arc $CA$ (which is longer than CE because it is not the shortest path). So we have at the very least $$\sin x \lt x$$ Now we need to show that line $BA=\tan x \gt x$.
Lines $AD$ and $CD$ are both tangent to arc $CA$. $CD+DA$ is longer than arc $CA$ because the set of points bound by sector $OCA$ is a subset of the set of points bound by quadrilateral $OCDA$, both of which are convex sets. This means that the perimeter of quadrilateral $OCDA$ must be longer than the perimeter of the sector $OCA$ (as per Archimedes, On the Sphere and Cylinder Book I). But both the sector and the quadrilateral both have sides $OC$ and $OA$, so we have $$CA=x<DC+DA$$ But $BD>CD$ because it is the hypotenuse in $\triangle BCD$ we have $$\tan x = BA = BD+DA\gt CD+DA \gt CA=x \gt \sin x$$

So we have $$\sin x \lt x \lt \tan x$$ $$\frac{\sin x}{x} \lt 1 \lt \frac{\tan x}{x}=\frac{\sin x}{x}\cdot\sec x$$ From this we can extract $$\frac{\sin x}{x} \lt 1$$ and $$1 \lt \frac{\sin x}{x}\cdot\sec x$$ $$\cos x \lt \frac{\sin x}{x}$$ Putting these inequalities back together we have $$\cos x \lt \frac{\sin x}{x} \lt 1$$

Because $\displaystyle\lim_{x\to 0}\cos x = 1$, by the squeeze theorem we have $$\lim_{x\to 0}\frac{\sin x}{x}=1$$

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The reason that I chose an arc length proof is because most derivations that I've seen of the area of a circle (see assume that $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$. Using an area based proof, to me, seems like putting the cart before the horse. – John Joy Nov 9 '14 at 15:14
How did Archimedes avoid assuming $\frac{chord}{arc}\rightarrow 1$ as $arc$ tends to zero in his argument about perimeters that you referred to? I asked the following question because this issue has kept haunting me for a while. – String Feb 5 '15 at 9:34
Actually, what Archimedes assumed was that if 2 curves (both concave in the same direction) have the same endpoints and one of the curves is strictly between a line segment with the same end points and the other curve, then the length of the enclosed curve is shorter than the length of enclosing curve. See axiom #2 – John Joy Feb 5 '15 at 14:17
OK, so he essentially axiomatizes the idea that among all curves connecting endpoints $A$ and $B$ satisfying some kind of regularity condition - concavity in the same direction, we have this inequality based on enclosure. This makes $arc(CA)<CD+DA$ in your figure merely equivalent to the assumption itself, but it makes sense and is very nicely put by Archimedes! Thank you for pointing me to that! – String Feb 5 '15 at 14:52

Here's one more: $$ \lim_{x \to 0} \frac{\sin x}{x}=\lim_{x \to 0} \lim_{v \to 0}\frac{\sin (x+v)-\sin v}{x}\\ =\lim_{v \to 0} \lim_{x \to 0}\frac{\sin (x+v)-\sin v}{x}=\lim_{v \to 0}\sin'v=\lim_{v\ \to 0} \cos v=1 $$

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Usually, this limit is used to compute the derivative of $\sin(x)$. – robjohn Jul 20 '13 at 18:43

It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.

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Don't you feel strange about why most of the proofs are done with a figure? I've had this problem in the beginning, and realized after that this is due to the definition we use for the function $\sin x$. Because the usual definition of $\sin x$ we all study first in high schools depends on “classical geometry” and usually with a figure, you should depict out the figure and to make it clear.

However, if you use other definitions of $\sin x$ that are equivalent to the former, you'll find it more simple. For example,

$$\sin x = \frac{x^1}{1!} - \frac{x^3}{3!}+ \frac{x^5}{5!} - \cdots + \cdots - \cdots$$

and hence

$$\frac{\sin x}x = \frac{x^0}{1!} - \frac{x^2}{3!}+ \frac{x^4}{5!} - \cdots$$

which obviously tends to $1$ as $x$ approaches 0.

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Indeed, it's easy to see that this holds if one uses a series, but this question starts on the prerequisite that one does not use a series. – FUZxxl Mar 13 '15 at 17:54
And how do you know that the derivative of an infinte series is equal to the sum of the derivatives of each term? – Steven Gregory Apr 16 at 4:32

Simple one is using sandwich theorem Which demonstrated earlier.In this method you have to show that $\frac{\sin x}{x} $ lies between other two functions. As $x \longrightarrow 0$ both of them will tends to ONE.

Then as in the case of sandwich (if both the bread part go to one stomach the middle portion will also go to the same stomach) $\frac{\sin x}{x}$ will go to ONE.

You can use geogebra to see the visualization of this phenomena using geogebra.First you input $\sin x$ and $x$ and observe that near to $0$ values of $\sin x$ and $x$ are same.

Secondly input $\frac{\sin x}{x}$ then observe function is approaching to $1$ as $x$ tends to $0$

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This describes the Sandwich Theorem, but does not answer the question. At best, this should be a comment to the question, – robjohn Dec 13 '14 at 13:31

An approach that hasn't been mentioned is finding $\frac{d\arcsin y}{dy}$ first. This is easy because geometrically when interpreting $\arcsin$ as an area, $\arcsin y = \int_{0}^y \sqrt{1-Y^2} dY - \frac{1}{2}y\sqrt{1-y^2}.$ Differentiating with respect to $y$ gives $\frac{d\arcsin y}{dy} = \frac{1}{\sqrt{1-y^2}}$. Using the theorem for the derivative of inverse functions yields $\sin' \theta = \sqrt{1 - \sin^2 \theta} = \cos \theta$. So $\sin' 0 = 1$.

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While this is indeed an interesting approach, integrals haven't been taught at the point where this limit is proved. Thank you for your answer though. – FUZxxl Apr 11 '15 at 17:38

Because $\sin x$ has zeroes at $x=n \pi$ for arbitrary integer $n$ including $x=0$, you can use Vieta's Theorem $\sin x = A(\cdots(x+2 \pi)(x+\pi)x(x-\pi)(x-2 \pi)\cdots)$ with a constant $A$. Because $\sin(\frac{\pi}{2})=1$ this constant can be determined by the equation: $$1=A(\cdots(\frac{\pi}{2}+2 \pi)(\frac{\pi}{2}+\pi)\frac{\pi}{2}(\frac{\pi}{2}-\pi)(\frac{\pi}{2}-2 \pi)\cdots).$$

Now, in the Expression $f(x):= \frac{\sin(x)}{x}$ the $x$ cancels such that $$f(x)=A(\cdots(x+2 \pi)(x+\pi)(x-\pi)(x-2 \pi)\cdots),$$ hence: $$\lim_{x \rightarrow 0} f(x)=A(\cdots(2 \pi) \cdot \pi\cdot(- \pi)\cdot(-2 \pi)\cdots) = A \prod_{k=1}^\infty (-k^2 \pi^2).$$

$\frac{1}{A} = \frac{\pi}{2} \prod_{k=1}^\infty ((\frac{\pi}{2})^2-k^2 \pi^2)$. The proof is completed when the Wallis product is used.

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None of the infinite products in your answer converge. – Antonio Vargas Oct 15 '15 at 1:30
Ontop of Antonio Vargas comment, Vieta's theorem only works for finite number of zeros. You need the Weierstrass-Factorization Theorem or Hadamard-Factorization Theorem. Btw you also need to verify that $\sin(x)$ does not have any complex zeros. – MrYouMath Jun 10 at 12:52

Simple Proof:
$$\lim\limits_{x\to 0}\frac{\sin x}x=1$$
it takes $0/0$ form,
so using L'Hôpital's rule, $$\lim\limits_{x\to 0}\frac{\sin x}x=\lim\limits_{x\to 0}\frac{\cos x}1=1$$

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It's nice that you post an answer that was literally posted as the top comment to this question. – FUZxxl Oct 20 '13 at 22:15
No one reads comments it seems – Squirtle Dec 23 '13 at 19:45
Be careful!! This proof is circular. See this meta answer, and also this comment. If you know that the derivative of $\boldsymbol{\sin x}$ is $\boldsymbol{\cos x}$, then you have already at some point proven that the limit is $\boldsymbol{1}$. – 6005 May 2 '14 at 16:42
Also there's nothing wrong with posting an answer identical to a comment. Comments such as that should be made into answers so that they can be properly discussed and voted on. – 6005 May 2 '14 at 16:44

Just for variety, I am going to include an $\epsilon$-$\delta$ proof:

Using the epsilon delta definition (am going to use it in for $\lim_{y\to 0} {\sin y\over y} = l, l \in \mathbb R$) gives the following:

Given a $\epsilon > 0 \exists \delta$ such that $|{\sin y\over y} - l| < \epsilon$ for $y \in [y_0-\delta, y_0+\delta], l \in \mathbb R$

Breaking up the absolute value gives:

$$l-\epsilon < {\sin y\over y} < l+\epsilon$$

multiply both sides by y giving:

$$y(l-\epsilon) < \sin y < y(l+\epsilon) =$$

$$yl-y\epsilon< \sin y < yl + y\epsilon$$ Since $l \in \mathbb R$ we have a new $\epsilon_1 = {\epsilon\over l}$ The inequality becomes $$y-y\epsilon_1< \sin y < y + y\epsilon_1$$ Then divide by $y \in \mathbb R$ $$ {y-y\epsilon_1\over y}< {\sin y\over y} < {y + y\epsilon_1\over y} $$ Then $$ 1-{y\epsilon_1\over y}< {\sin y\over y} < 1 + {y\epsilon_1\over y} $$

Then rewrite as the definition of limit: $$ |{\sin y\over y} - 1| < \epsilon_1 $$

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How did you obtain $y-y\epsilon_1<\sin y<y+y\epsilon_1?$ It should be $y-y\epsilon_1<\frac{\sin y}{\ell}<y+y\epsilon_1$ instead. – Singh Mar 11 at 7:22
Where you use the fact that you have a sin function in this proof ??? – Hamza Jun 15 at 1:33

$y=x$ is the tangent to $f(x)=sinx$ at $x=0$, so in the neighbourhood of zero, the graph of $sinx$ follows graph of $y=x$ hence $$\lim_{x \to 0} \frac{sinx}{x} \to 1$$ Correct me if this is wrong Intuitively

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How do you prove that $y=x$ is that tangent? – FUZxxl Apr 29 at 15:05

Here is a geometric proof of the formula: $$\frac{\sin(x)}{x} = \frac{O}{H}/\frac{a}{r}=\frac{O\cdot r}{H\cdot a}$$ where $a/r$ is the angle $x$ in radians with $a$ as the arc length and $r$ as the radius. But $H=r$ so, if we call them both $hr$: $$=\frac{O\cdot hr}{hr\cdot a}=\frac{O\cdot hr}{a\cdot hr}=\frac{O}{a}\times1$$ and we are setting both $O$ and $a$ to zero, so finally: $$=\frac{0}{0}\times1=1\times1=1$$ The most controversial step here is equating $0/0$ to $1$, but these zeros are not variables approaching zero - they are set as absolutely zero, so the rule $a/a=1$ should apply.

EDIT: this proof posits a right-angle triangle based on the $x$ axis with the hypoteneuse H as a radius; O is the opposite side (the adjacent A is not used). Also, the technique here is most similar to @MichaelHardy's approach, but made as explicit as possible.

EDIT2: the justification for setting $0/0\rightarrow1$ is that $O$ and $a$ are equal at zero.

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It might be better to make a second, self-answered question that asks to prove the equality using the Taylor series and then link further duplicates to that question instead of adding an off-topic answer here. – FUZxxl Jun 10 at 12:40
@FUZxxl Found the geometrical answer and totally rewrote the post. – selfawareuser Jun 14 at 9:36
@selfawareuser: What is $O$ and $H$? How are they defined? A diagram would probably help. I also do not follow the last part which you describe as being controversial. Finally, did you read the other answers and do you think this approach brings something to the table that cannot already be found in one of them? – String Jun 14 at 11:28
@String I've had debates about $0/0$ but I'm happy with it. I did read the other answers - I prefer this approach, it nicely finishes off another answer I posted. – selfawareuser Jun 14 at 13:03
@selfawareuser: Thanks for clarifying what is meant by $O,H$! That part now makes sense. But why $0/0$ should be equal to $1$ is still a mystery to me. The function $f(x,y)=x/y$ does not have a removable singularity at $(0,0)$ and I do not see why we should simply choose to apply the rule $f(a,a)=a/a=1$ and extend that to $a=0$. Link to Numberphile video with a discussion of this problem. – String Jun 14 at 14:21

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