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

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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 Oct 23 '11 at 22:51
+1, nice site - What is the length BC? –  Emmad Kareem Oct 24 '11 at 5:34

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

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

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

(This answer is actually an answer to this question, but I post it here since that question has been closed. If that question is reopened, I will move my answer there.)

I believe the difficulty many students have in understanding the proof of $$\lim_{x\to0}\frac{\sin x}{x}=1$$ is that it often comes too early in a calculus course, before one has learned integration, and therefore before one has learned how to think about arc lengths and areas. Since arc length (or area) enters fundamentally into the definition of the $\sin$ function, the proofs that are usually given end up relying on geometric arguments that seem out of place next to the style of argumentation that one otherwise encounters in the first part of a calculus course. I'm not suggesting that these arguments aren't rigorous, or that there aren't good reasons for bringing up trigonometric functions early—but I believe it can be satisfying to understand how this result comes about, rather straightforwardly, once one has defined the trigonometric functions in calculus terms.

If definite integration and arc length are unfamiliar to you, then this proof may not help. However, the notion of arc length is used only to motivate a definition, and all one needs to know about the definite integral is that it represents the area under the curve. This is used to obtain upper and lower bounds.

We make a purely algebraic definition of the $\arcsin$ function as follows. For $y\in(-1,1)$ let $$\arcsin y=\int_0^y\frac{dt}{\sqrt{1-t^2}}.$$ Interpreted geometrically, this is, in fact, the formula for the signed length of the arc of the unit circle that extends from $(1,0)$ to $(\sqrt{1-y^2},y)$, and therefore indeed the $\arcsin$ function as usually defined, but this definition itself involves no geometry. The $\sin$ function is then the inverse of this function, at least in an interval containing 0. (In fact, in the interval $(-\pi/2,\pi/2),$ where the symbol $\pi$ represents twice the upper bound of the $\arcsin$ function. Thinking about how to extend the function to the entire real line is not needed for our purpose.)

We will first show that $$\lim_{y\to0}\frac{\arcsin y}{y}=1.$$ Since $\sqrt{1-t^2}$ is bounded above by $1,$ the integrand defining $\arcsin y$ is bounded below by $1,$ and, for $y$ positive, $\arcsin y$ is bounded below by $y.$ Hence $(\arcsin y)/y$ is bounded below by $1.$ Since, in the integral, $\sqrt{1-t^2}$ is bounded below by $\sqrt{1-y^2},$ the integrand is bounded above by $1/\sqrt{1-y^2},$ and, for $y$ positive, $\arcsin y$ is bounded above by $y/\sqrt{1-y^2}.$ Hence $(\arcsin y)/y$ is bounded above by $1/\sqrt{1-y^2}.$ Since $(\arcsin y)/y$ is an even function, the same bounds apply for $y$ negative. But both of these bounds have limit $1$ as $y\to0,$ and the result follows from the squeeze theorem.

This is almost the result we want. To finish, define the continuous function $$f(y)=\begin{cases}\frac{\arcsin y}{y} & \text{for y\in(-1,1)\setminus\{0\},}\\ 1 & \text{for y=0,}\end{cases}$$ by plugging the hole at $y=0$ using the limit we just computed. Let $g(x)=\sin x.$ Since $f$ and $g$ are continuous at $0$ and $g(0)=0,$ we have $$\lim_{x\to0}f(g(x))=f\left(\lim_{x\to0} g(x)\right)=f(0)=1.$$ But for $x\ne0,$ we have $$f(g(x))=\frac{\arcsin\sin x}{\sin x}=\frac{x}{\sin x},$$ and so we have our result.

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My answer is really a non sequitur to the original post. (As noted, it's an answer to a slightly different question.) In particular, the goal of the original poster was to use the result to prove continuity of $\sin,$ whereas continuity is assumed in my answer. Continuity of $\arcsin$ on $(-1,1)$ follows in the definition I have used from continuity of the integrand and the Fundamental Theorem of Calculus (which actually implies differentiability). Continuity of $\sin$ follows from the easy fact that $\arcsin$ is... –  Will Orrick Jan 26 at 15:54
...one-to-one on $(-1,1)$ and from the theorem that the inverse of a continuous, one-to-one function is continuous. –  Will Orrick Jan 26 at 15:54
I do not like the proof via squeeze theorem that $\sin x < x < \tan x$ because how do you prove that $x<\tan x$? One can prove that $\sin x < x$ as the line is the shortest distance between two points. But what kind of geometric argument do you use to justify that $x<\tan x$? I do not see of one. So I believe it is not possible to prove this limit using lengths. Instead one has to apply the squeeze theorem using areas, as it standard in calculus textbooks.
But then a question comes up. Why is the area of a sector of a unit circle equal to $\tfrac{1}{2}\theta^2$? This can be justified without the use of calculus as there many ways of deriving areas and lengths of sectors for circles without using anything about derivatives or integrals of the trigonometric functions.