# How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

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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I assume you mean 0, not infinity? –  mixedmath Oct 23 '11 at 16:24
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
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

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
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
@Mike: I agree that there is a bit of faith that goes into the mix at some early stages, but I think this is one that a little bit of hand-waving can make believable. –  robjohn Oct 23 '11 at 23:40
Nice and coloured! :) (+1) –  Chris's sis Aug 23 '12 at 22:38

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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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.

$$\lim_{x\rightarrow0}\frac{\pi\sin(x)}{x(1+\sin(x))}={\pi}\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x(1+\sin(x))}=1\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$

The proof is complete.

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http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.html

Here is the picture copied from that blog:

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@robjohn: see, I like the picture under this link much more because it uses $\theta$ as an arclength instead of using $\theta/2$ as an area. –  Mike F Oct 23 '11 at 22:51
+1, nice site - What is the length BC? –  Emmad Kareem Oct 24 '11 at 5:34
+1 same idea as my proof, but much cleaner. –  John Joy Nov 9 at 15:07

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

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 $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$ –  metacompactness Jul 20 '13 at 19:15

Simple Proof
$$\lim\limits_{x\to 0}\frac{\sin x}x=1$$
it takes $0/0$ form,
so using L'Hospital 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 $\boldsymbol{1}$. –  Goos May 2 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. –  Goos May 2 at 16:44

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

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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

I claim that for $0<x<\pi/2$ that the following holds $$\sin x \lt x \lt \tan x$$
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 mathopenref.com/circleareaderive.html) 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 at 15:14

Here's a simple explanation:

$Sinx$ is opposite side by hypotenuse $x$ is arc length by radius.

When $x$ becomes smaller and smaller opposite side tends to be equal to arc lenghth and hypotenuse tends to radius and hence the ratio tends to 1

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This is not a proof - just an intuition! –  pushpen.paul Nov 9 at 7:18

Simple one is using sandwich theorem Which demonstrated earlier.In this method you have to show that $\frac{sinx}{x}$ 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) $\frac{sinx}{x}$ will go to ONE.

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

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

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$\\ \sin x=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}\\ \implies\dfrac{\sin x}{x}=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n+1)!}\\ \implies \displaystyle\lim_{x\to 0} \dfrac{\sin x}{x}=\displaystyle\lim_{x\to 0} \displaystyle\sum_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n+1)!}\\ \implies \displaystyle\lim_{x\to 0} \dfrac{\sin x}{x}= \displaystyle\sum_{n=0}^\infty\left(\displaystyle\lim_{x\to 0}\dfrac{(-1)^nx^{2n}}{(2n+1)!}\right)\\ \implies \displaystyle\lim_{x\to 0} \dfrac{\sin x}{x}=1$

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This answer uses the series expansion of $\sin.$ The question specifically states to not use this series. –  FUZxxl Nov 12 at 20:49
@FUZxxl: Yes, I have noted that. But in higher mathematics $\sin$ is not defined geometrically. It is by the series. Using geometrical definition of $\sin$ may lead to some non-rigorous proofs. This is the reason I wrote this answer. –  user 170039 Nov 13 at 3:33
You did misunderstand the question then. I wrote this question back when I was still in school with the specific explanation that we defined $\sin$ geometrically and that we cannot use the Taylor series for we haven't proved that it's equal to the geometric definition. –  FUZxxl Nov 13 at 10:31
@FUZxxl: You may see this. –  user 170039 Nov 13 at 11:47