I'm having difficulty understanding a proof of $\lim_{x\to0}\frac{\sin{x}}{x}=1$ provided in Simmons's Calculus With Analytic Geometry, pg. 72. The proof goes as follows:

Let $P$ and $Q$ be two nearby points on a unit circle, and let $\overline{PQ}$ and $\widehat{PQ}$ denote the lengths of the chord and the arc connecting these points. Then the ratio of the chord length to the arc length evidently approaches 1 as the two points move together:

$\frac{\text{chord lenght}\overline{PQ}}{\text{arc lenght}\widehat{PQ}}\to1$ as $\widehat{PQ}\to0$

With the notion in the figure, this geometric statement is equivalent to

$\frac{2\sin{\theta}}{2\theta}=\frac{\sin{\theta}}{\theta}\to1$ as $\theta\to0$


My doubt is, doesn't this proof simply that $\lim_{x\to0}\frac{\sin{x}}{x}=\frac{0}{0}$? I mean, sure the ratio of $\text{chord lenght}\;\overline{PQ}$ to $\text{arc lenght}\;\widehat{PQ}$ approaches 1 as $\theta$ approaches $0$, but that's because both the numerator and the denominator approach the samue value, which is $0$. How is it any different from saying $\lim_{x\to0}\frac{\sin{x}}{x}=1$ because both $\sin{x}$ and $x$ approach the samue value $0$ as $x\to0$?

  • 6
    $\begingroup$ Actually, this proof contains an act of faith, based on a geometric intuition, that the chord length gets closer and closer to the arc length as the arc length gets smaller and smaller. $\endgroup$
    – Bernard
    Commented Feb 15, 2017 at 16:25
  • $\begingroup$ if you've got two circles, and one circle is much larger than the other, and you take a set length of the curve, say 10cm - on the large curve, this corresponds to a much smaller $\theta$ - I think you can imagine, and prove mathematically that the large circle portion is closer to a straight line, hence $2\theta$ is becoming closer to 2sin(theta) - but that is without them having to become zero $\endgroup$
    – Cato
    Commented Feb 15, 2017 at 16:30
  • $\begingroup$ In fairness to the late Dr. Simmons, the "proof" on p72 is a sketch of a proof, the second such sketch on the page, intended to establish face validity of the limit so that he can move forward with a preliminary discussion of derivatives of elementary functions. Real proofs of the limit begin in Chapter 9, the chapter on trigonometric functions. I count 4 demonstrations of this limit in the text, including the "usual" one mentioned below. This layered approach to establishing proofs is characteristic of his writing. $\endgroup$ Commented Mar 27 at 12:56

5 Answers 5


I think your problem is deeper than understanding this proof.

What makes calculus difficult is that it's hard (at first) to make sense of the ratio of two small quantities. Just because both "approach the same value $0$" doesn't mean the limit of the quotient is $1$.

For example $$ \frac{2x}{x} $$ clearly approaches $2$, since it's always exactly $2$ when $x$ is not $0$.

Also $$ \frac{2h+h^2}{h} = 2 + h $$ approaches $2$ as $h$ approaches $0$, even though it looks like $0/0$ when $h = 0$. That is in fact just the limit you probably saw when finding the slope of a parabola (differentiating $y=x^2$ when $x=1$).

The moral of the story is that you have to pay attention to the relative rates at which the numerator and denominator approach $0$. In the question you ask, the rates are the same and the limit is $1$.

You can take comfort in the fact that this was a mathematical and philosophical difficulty when Newton and Leibniz were inventing calculus.

Note: this in an informal discussion. The formal definition of limits avoids words like "approaches". And @Bernard 's comment about the geometric argument for $\sin(x)/x$ correctly notes the informality there too.

  • 1
    $\begingroup$ I believe this is most instructive answer for OP. I would, perhaps, add more emphasis on how many many limits high school and university students do are indeed of the form $0/0$, yet limits themselves can be any number, not just $1$ (this is what you said, I would just insist more on this point). $\endgroup$
    – Ennar
    Commented Feb 15, 2017 at 20:27

I like this figure better. enter image description here

There are two triangles here and one section of a circle with an angle of measure x.

The base of the two triangles is 1. The heights are $\sin x, \tan x$

The areas are: smaller triangle $\frac 12 \sin x,$ section of the circle $\frac 12 x,$ larger triangle $\frac 12 \tan x.$

$|\sin x|\le |x| \le |\tan x|\\ 1\le \frac {x}{\sin x} \le \sec x\\ 1\ge \frac {\sin x}{x} \ge \cos x$

Let $x$ go to $0$ and $\frac {\sin x}{x}$ gets squeezed.


The questions raised in the last paragraph of your post are best handled by the excellent answer from Ethan Bolker. The following is a feedback on the proof given in your book.

I really can't believe that this is an actual proof given in an actual book. If this is really the case as you say, then this is one of the boldest examples of intellectual dishonesty perpetrated by most calculus textbook authors.

Note that $$\lim_{x\to 0}\frac{\sin x} {x} = 1\Rightarrow \frac{\text{chord length } PQ} {\text{arc length } PQ} \to 1 \text{ as arc length }PQ\to 0$$ and not the other way round as the proof in your book seems to suggest. For the case of a circle, the ratio of chord length to arc length tends to $1$ but this is not as self evident as Simmons writes, rather it depends on the limit in question here. The proof presented in the book is just plain hand-waving.

The proof for the limit can be easily given using areas of sectors instead of lengths of arcs because the analysis of arc length is significantly more cumbersome compared to that of areas. The standard proof is the one mentioned by user robjohn in this answer.

When using arc length we have two options:

  • Establish that the ratio of chord length to arc length tends to $1$ as the arc length tends to $0$ for a general class of sufficiently well behaved curves (and circle is one such curve) and then argue as in your book. This can be done by having a rigorous analysis of length of arc of a general curve and using some theorems from integral calculus.

  • Treat the case of circle on special basis and establish the inequality $\sin x<x<\tan x$ for $x\in(0,\pi/2)$ using length of arc of a circle and then apply squeeze theorem.


A more correct geometric proof would consist in considering the intersection $P'$ and $Q'$ of the lines $(OP)$ and $(OQ)$ with the tangent to the circle at the midpoint $A$ of the arc $PQ$. The triangle OPQ is contained in the circle sector OPQ, which is itself contained in the triangle $OP'Q'$. Hence we have the corresponding inequalities between the areas of these domains (we suppose the circle has radius $1$): $$\sin\theta\cos\theta\le \theta\le\tan\theta=\frac{\sin\theta}{\cos\theta}.$$ The l.h.s. inequality is equivalent to $$\frac{\sin\theta}\theta\le\frac1{\cos\theta},$$ and the r.h.s. inequality to $$\frac{\sin\theta}\theta\ge\cos\theta$$ As $\cos\theta$ is continuous at $0$, the squeezing principle yields the limit you were longing for.


One can interpret the proof in a more abstract, intuitive way. As arc length decreases, its "curviness" disappears (i.e. it becomes a straight line). We can say this as $\sin(\theta) \to \theta $ as $ \theta \to 0$. Taking this into account:

$$\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = \lim_{\theta \to 0} \frac{\theta}{\theta} = \lim_{\theta \to 0} 1 = 1$$

  • 4
    $\begingroup$ Careful. You probably used this limit to calculate the derivative of $\sin$. In that case using L'Hopital is circular reasoning. $\endgroup$ Commented Feb 15, 2017 at 16:35
  • $\begingroup$ @AnthonyD'Arienzo I think it is a very poor example of L'Hospital's rule, as Ethan pointed out... $\endgroup$
    – imranfat
    Commented Feb 15, 2017 at 16:47
  • $\begingroup$ You're right, good catch. $\endgroup$ Commented Feb 15, 2017 at 19:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .