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I'm looking for a convincing proof, using first principles, that $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ (Please use ordinary unit circle definitions of trigonometric functions.)

It occurred to me that the classic proof, which compares three areas, uses the formula ${1\over 2}r^2\theta$ for the area of a circular sector of angle $\theta$, which in turn assumes the area of a circle is $\pi r^2$. But this fact is almost always proven in texts using an integral, which ends up using the derivatives of $\sin$ and $\cos$, and we're back to that limit again.

So I need a non-circular proof that doesn't rely on playing definition games ("let $\sin$ be the following power series..."). The answer to this question is definitely playing definition games.

Sorry for the pun.

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    The formula for area does not require anything regarding sine and cosine; it suffices to have the perimeter formula $P = 2 \pi r$, which comes from the definition of $\pi$. – Ben Grossmann Dec 08 '15 at 14:22
  • @Arthur but how do you know that $x < \tan(x)$? – Ben Grossmann Dec 08 '15 at 14:28
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    It's tricky, because at heart, you need to know some deep properties about paths in the plane. For example, the distance $\sin x$ is the length of shortest path from $(\cos x,\sin x)$ to the $x$-axis, and the length $x$ is the length of a circle arc from $(\cos x,\sin x)$ to the $x$-axis, so we know $\sin x\leq x$. Then you can show with a similar, but more complicated theorem about paths, a bound that $x\leq \tan x$. Then show that $\cos x\to 1$. – Thomas Andrews Dec 08 '15 at 14:29
  • I honestly do not think there is another way to get that limit other than the "staple-method" area inside circle - circle - area outside circle approach. I have reviewed many textbooks on elementary Calculus for our institution and always looked for a different approach on this topic. Nope, never found it. Remember though that the standard area method is generally easy "accessible" for students at this level. If there is another method that does not involve "circular logic", I would really like to know... – imranfat Dec 08 '15 at 14:44
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    I don't think Archimedes used integrals for computing the area of circles and circular sectors. – egreg Dec 08 '15 at 14:48
  • I'm sure there is a way to prove the area of a sector is $\frac{1}{2} r^2 \theta$ from within Euclidean geometry. And once you have that everything is straightforward (you draw the triangle inside the sector inside the other triangle and compare areas). – Ian Dec 08 '15 at 14:50
  • A way to start the proof (semi-formally): the circle is invariant under rotation. Therefore the area of a sector is the area of the circle, times the length of the arc swept out by the sector, divided by the circumference of the circle. Now by the definition of the radian, $r \theta$ is the length of the arc swept out. So now we need $\frac{1}{2} r^2 \theta = \frac{A r \theta}{C}$, i.e. we need $\frac{A}{C}=\frac{r}{2}$. – Ian Dec 08 '15 at 14:55
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    Hint: the area of the unit circle approximated by a regular $n$-gon is $\sin(2\pi/n)n/2$. –  Dec 08 '15 at 14:56
  • We can find the formula for the area of a circle by arguing that $\frac {d}{dr}A(r) = P(r) = 2 \pi r$. – Ben Grossmann Dec 08 '15 at 15:03
  • The analysis approach - defining $\sin x$ as a power series, then proving it has a ton of properties, and finally that it must be your usual $\sin x$ - is perfectly non-circular. If it makes you happier, call the power series $s(x)$ and then prove that $s(x)=\sin(x)$. It's not the same as a the type of proof you are seeking, but it is non-circular. – Thomas Andrews Dec 08 '15 at 15:16
  • @ThomasAndrews: How can one prove from the power series of $\sin$ that it "must be" the "usual $\sin x$"? –  Dec 08 '15 at 15:24
  • That's the fun part, but it is a perfectly standard argument in analysis. Requires a few pages of work, though, so I'm not putting it here in comments. @user170039 – Thomas Andrews Dec 08 '15 at 15:25
  • @ThomasAndrews: Can you then site some reference(s)? –  Dec 08 '15 at 15:26
  • @ThomasAndrews: Does the argument also entail the result that $\sin x$ is the ratio $\frac{\text{perpendicular}}{\text{hypotenuse}}$? –  Dec 08 '15 at 15:29
  • Off the top of my head, no. I've seen it done it books, but I'm away from my books. It's a fairly standard argument. Usually best to start proving stuff about $\exp(z)$, then move from there. And yes, you ultimately prove that $s(x)$ is the same as that ratio of perpendicular to hypotenuse when $x$ is the length of the curve on the circle of the same angle. @user170039 – Thomas Andrews Dec 08 '15 at 15:29
  • One can use existence and uniqueness theorem of differential equations to show that $s(x)=\sin x$. –  Dec 09 '15 at 09:08
  • see similar question http://math.stackexchange.com/q/1097757/72031 – Paramanand Singh Dec 10 '15 at 06:19

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I don't see anything circular in comparing areas to get the inequality $\sin x < x < \tan x$ for $0 < x < \pi/2$. However we need to be very cautious in defining the symbols $\sin x, \tan x$ properly given $x$ a real number.

The approach based on areas goes like this. Using the concept of definite integrals it can be proven that a sector of a circle has an area. This does not require anything beyond the continuity of the function $\sqrt{1 - x^{2}}$ in interval $[0, 1]$. In particular justification of the area of a circle is not dependent on the definition of trigonometric functions and $\pi$.

Next consider a unit circle with origin $O$ as center and let $A$ be the point $(1, 0)$. Let $P$ be any point on the circle. For our purposes it is sufficient to consider $P$ to be in first quadrant. Let the area of sector $AOP$ be $y$ so that $y > 0$. Also let $x = 2y$ and then by definition the point $P$ is $(\cos x, \sin x)$. This is the usual definition of trigonometric functions as studied at the age of 15 years or so.

Note that some textbooks base the definition of $\sin x, \cos x$ on the basis of length of arc $AP$ which is $x$. The definition is equivalent to the one based on areas of sectors, but comparing areas of figures is simpler than comparing the length of arcs (at least in this context). Consider the tangent $AT$ to unit circle at point $A$ such that $OPT$ is a line segment. Also let $PB$ be a perpendicular to $OA$ and $B$ is the foot of this perpendicular. Now it is easy to show that $$\text {area of }\Delta AOP < \text{ area of sector }AOP < \text{ area of }\Delta AOT$$ (because each region is contained in the next). However it is very difficult to compare the length of arc $AP$ with the length of line segments $PB$ and $AT$ (because there is no containment here).

The above inequality leads to $$\sin x < x < \tan x$$ from which we get $\sin x \to 0$ as $x \to 0$ and then $\cos x = \sqrt{1 - \sin^{2}x} \to 1$. Further the inequality is equivalent to $$\cos x < \frac{\sin x}{x} < 1$$ and hence $(\sin x)/x \to 1$ as $x \to 0$.

Update: It appears from OP's comments that the relation between length of an arc of a circle and area of corresponding sector is something which can't be proven without using any analytic properties of circular functions. However this is not the case.

Let $P = (a, b)$ be a point on unit circle $x^{2} + y^{2} = 1$ and let $A = (1, 0)$. For simplicity let's consider $P$ in first quadrant so that $a, b$ are positive. Then the length of arc $AP$ is given by $$L = \int_{a}^{1}\sqrt{1 + y'^{2}}\,dx = \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ The area of the sector $AOP$ is given by $$A = \frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx$$ We need to prove that $L = 2A$. We will do this using the fact that $b = \sqrt{1 - a^{2}}$ and using integration by parts.

We have \begin{align} \int\sqrt{1 - x^{2}}\,dx &= x\sqrt{1 - x^{2}} - \int x\cdot\frac{-x}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \frac{1 - x^{2} - 1}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\,dx + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int\sqrt{1 - x^{2}}\,dx &= \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\int \frac{dx}{\sqrt{1 - x^{2}}}\notag\\ \end{align} Hence $$\int_{a}^{1}\sqrt{1 - x^{2}}\,dx = - \frac{a\sqrt{1 - a^{2}}}{2} + \frac{1}{2}\int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ or $$\int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = 2\left(\frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\right)$$ or $L = 2A$ which was to be proved.

Contrast the above proof of relation between length and area with the following totally non-rigorous proof. Let the length of arc $AP$ be $L$. Then the angle subtended by it at the center is also $L$ (definition of radian measure). Divide this angle into $n$ parts of measure $L/n$ each and then the area of sector $AOP$ is sum of areas of these $n$ sectors. If $n$ is large then area of each of these $n$ sectors can be approximated by area of the corresponding triangles and this area is $$\frac{1}{2}\sin (L/n)$$ so that the area of the whole sector $AOP$ is $(n/2)\sin(L/n)$. As $n \to \infty$ this becomes $L/2$ and here we need the analytic property of $\sin x$ namely $(\sin x)/x \to 1$ as $x \to 0$. Therefore area can't be the basis of a proof of this limit. This is perhaps the reason that proofs for limit formula $(\sin x)/x \to 1$ looks circular.

A proper proof can't be done without integrals as I have shown above. Hence the proof that $(\sin x)/x \to 1$ depends upon Riemann integration and definition of $\sin x, \cos x$ as inverses to the integrals. This is same as $e^{x}$ is defined as inverse to integral of $1/x$.

Also see my another answer to a similar question.

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Paramanand Singh
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  • I have actually never seen this definition in terms of areas of sectors. Interesting proof. – Ian Dec 09 '15 at 14:02
  • How would you find the area of a circle? –  Dec 09 '15 at 23:20
  • @NotNotLogical: Area of a unit circle by definition is $$4\int_{0}^{1}\sqrt{1 - x^{2}},dx$$ and if you may like, you can define $\pi$ by the equation $$\pi = 4\int_{0}^{1}\sqrt{1 - x^{2}},dx$$ – Paramanand Singh Dec 10 '15 at 03:52
  • @ParamanandSingh These are those definition games I was talking about! :) Take $\pi$ to be the ratio of a circle's circumference to its diameter. How would you evaluate that integral? If you make a trig substitution, how do you know what the bounds are? –  Dec 10 '15 at 03:54
  • @NotNotLogical: So your definition of $\pi$ is ratio of circumference to the diameter. Great! by your definition $$\pi = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ and clearly you know that $$2\int_{0}^{1}\sqrt{1 - x^{2}},dx = \int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ so your definition of $\pi$ is same as my definition of $\pi$. there is no need to use any trig substitution. – Paramanand Singh Dec 10 '15 at 04:01
  • @Ian: The definition based on arc length is common and perhaps more famous, but analytically more difficult to handle. I found the definition based on area in Hardy's classic "A Course of Pure Mathematics". – Paramanand Singh Dec 10 '15 at 04:03
  • @ParamanandSingh How do you get between those two integrals? –  Dec 10 '15 at 04:11
  • Let us continue this discussion in chat. – Paramanand Singh Dec 10 '15 at 04:12
  • @NotNotLogical: See updated answer. – Paramanand Singh Dec 10 '15 at 04:31
  • @ParamanandSingh Nice. I like your approach. I would do it a little differently, but the idea that I care about is that the area can be derived in terms of circumference. –  Dec 10 '15 at 14:29
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You show that $x<\tan x$ when $0<x<\pi/2$ by taking the path:

$$(\cos 2x,\sin 2x),(1,\tan x),(1,0)$$

This path, aside from the endpoints, is outside the unit disk. The two segments are tangent to the circle. There is a theorem which says that therefore it has to be longer than the shortest arc on the boundary of the unit disk. I asked a question about this a while back - it is a result of the Hahn Banach theorem.

So this path, of length $2\tan x$, is greater than or equal to the path on the circle, $2x$.

(That $\sin x\leq x$ is due to a simpler rule - the shortest path from a point to a line, is the perpendicular from the point to the line.)

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Thomas Andrews
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  • The Hahn Banach theorem certainly isn't needed in its full capacity. What you really need is a line in $\Bbb R^2$ separating a point from a circle; perhaps this can be justified with good ol' Euclidean geometry. – Ben Grossmann Dec 08 '15 at 14:39
  • Don't remember enough Euclidean geometry, though. It's certainly possible, I don't even know how Euclidean geometry deals with arc lengths rigorously. – Thomas Andrews Dec 08 '15 at 14:41
  • That's a fair point... not sure how to do arc lengths myself. – Ben Grossmann Dec 08 '15 at 14:44
  • Symmetry! This is a side-angle-side argument. If those points are $A,B,C$, respectively, then the triangles $COB$ and $AOB$ are congruent because the two sides are equal and the angle is equal. @Omnomnomnom – Thomas Andrews Dec 08 '15 at 14:53
  • A nice thing in this picture is that the line segments lie tangent to the circle. – Ben Grossmann Dec 08 '15 at 14:53
  • @TA Just saw it! Thanks – Ben Grossmann Dec 08 '15 at 14:54
  • Right (the tangents) which you only actually know if you know the symmetry, too, because it is not obvious the first segment is tangent. If the segments weren't tangent, then they'd partly go inside the disc, and the argument falls apart. – Thomas Andrews Dec 08 '15 at 14:54
  • Perhaps you can say that the arc is shorter than the line segments by an Archimedian limiting process. In particular, take a "short cut" to make your two line segments into $3$, then add more sides to get a shorter path still which is closer to the circle, then claim that the length of the circle is the infimum of all such path lengths. – Ben Grossmann Dec 08 '15 at 14:59
  • "Claim the length of the circle..." Well, that's sort of the question - why is it the infinimum? That's not usually how we define arc length. That assumes what we are trying to prove if we define the arc length as the infimum of all paths outside the circle. :) @Omnomnomnom First principles are hard. – Thomas Andrews Dec 08 '15 at 15:03
  • any computation of $\pi$ based on its definition requires some definition of arclength, which is ultimately a limit. It suffices then, to show that this limit is equal to whatever limit the arclength of a circle is defined to be. – Ben Grossmann Dec 08 '15 at 15:07
  • But the limit is not, in any definition, an infimum. That is pecular to curves on a convex set's boundary, and paths that are external to that convex set. @Omnomnomnom Indeed, every definition of arc length uses supremums rather than limits, and definitely not infimums. Infimums are defined when talking about geodesics - shortest paths between $a,b$. So you are basically just assume the circle arc is a geodesic, not proving it is. – Thomas Andrews Dec 08 '15 at 15:08
  • Fair point. First principles are hard. – Ben Grossmann Dec 08 '15 at 15:23
  • @ThomasAndrews Thanks, I like this approach. I think it gets to the heart of the matter. I was hoping for something simple, but probably there's no way around getting into analysis. –  Dec 08 '15 at 22:15
  • Interestingly, once you know this limit, you can easily show $(1+ix/n)^n$ converges to our expected $\cos x+i\sin x$. I'll link to that when I get home, if interested. – Thomas Andrews Dec 08 '15 at 22:22
  • Basically, $\sin x/x\to 1$ and the easily derivable $(\cos x-1)/x\to 0$ means that for fixed $x$, $\cos(x/n)+i\sin(x/n) = 1+\frac{ix}{n} + o(1/n)$, and therefore, by the argument at the linked answer $\cos x+i\sin x=\lim (1+ix/n)^n$. http://math.stackexchange.com/a/1451245/7933 – Thomas Andrews Dec 08 '15 at 23:24
  • The approach which you mention to prove $x < \tan x$ is justified by a much simple theorem dealing with convex paths. See http://math.stackexchange.com/q/1785841/72031 – Paramanand Singh May 15 '16 at 10:24
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You can also note that (for small, positive $x$) we have $$\sin(x)<x<\tan(x)$$then divide through by $\sin(x)$ and squeeze. This gives the limit of $\frac{x}{\sin x}$.

Arthur
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$\sin x$ is easy to define, but not $x$! You won't find a more elementary rigorous definition of the $\sin$ function other that is the inverse of the $\sin^{-1}$ function, where $\sin^{-1} y$ is defined for not too large $y$ as the length of the arc in the circle between $(1, 0)$ and $(\sqrt{(1-y^2)}, y)$. This is just the completely elementary definition with the units for the angle actually defined. It is not elementary enough for a first course since it uses inverse functions. The existence of the arc length can be hand-waved and the coordinates not stated in a first course.

Rigorously defining arc lengths needs a limit operation. Fairly elementary calculus gives the formula for the arc length of a circle: $$\sin^{-1} y = \int ^y _0 \frac{dt}{\sqrt{(1-t^2)}}.$$ Note that this is not circular; it is the definition of $\sin^{-1}$ and thus of $\sin$.

The value of the limit is obvious from this formula using elementary calculus. Note that I haven't done anything complicated like evaluating the integral. Other properties of the $\sin$ function can be derived, not so easily, from this definition. $\pi$ is by definition the length of a half-circle, and thus twice the value of this integral from $0$ to $1$, whatever that may be. Since the integral becomes improper at $y = 1$, better define $\pi$ using half of the first quadrant instead of all of it to minimize the calculus.

This is easier, at least for analysts, to understand after studying elliptic functions and elliptic integrals. Change $t^2$ to $t^3$ in the integral and the integral becomes elliptic. There is no geometry of the circle or arc length to help or confuse you in understanding this integral. The inverse of the elliptic integral is an elliptic (doubly periodic) function. This is much less easy to see and prove rigorously from the integral than that $sin$ is singly periodic. Other properties of elliptic functions are even harder to see and prove starting from the integral.

Bruce Evans
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  • Try not to use exclamation marks after variable names if you don't mean factorials. If there is any reason to exclaim here, it is unclear, except that what you are saying is deliberately confusing. Of course, $x$ is easy to define, and what you mean is that $\sin x$ is hard to define because defining the arc of length $x$ on the circle is hard. – Thomas Andrews Dec 08 '15 at 16:51
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    The geometric significance of the "$x$" in "$\sin(x)$" is what is difficult to define. – Ian Dec 08 '15 at 18:12