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I know one proof of $\lim_{x \to 0}\frac{\sin(x)}{x}=1$, the one using squeeze theorem. I have been told, however that this proof is not fully rigorous, as it uses the concept of arc length very informally. I wonder if any can provide me with a more rigorous proof of this fact. There are, of course, a several ways how to define the sine function. I don't care which one is used.

Potato
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Adam
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Since you don't care which definition I use: let sine be the function satisfying $$ y'' + y = 0~~~\text{with}~~~\left\{\begin{array}{l} y'(0) = 1\\y(0) = 0 \end{array}\right. $$ Then, by L'Hoptital's Rule, we have $$ \lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{1}{1} = 1 $$

Alec
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  • Is there a way to see that sine defined this way has all the familiar properties? – Adam Dec 15 '13 at 03:58
  • Yes, look it up! I'll see on my side if I can come up with something elegant enough to fit in a comment. – Alec Dec 15 '13 at 04:03
  • I guess the easiest way to look at this, if you are willing to accept a complex argument, is that from this DE definition, we $\sin(x) = \frac{1}{2}(e^{ix} - e^{-ix})$ and $\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$. From there, we get most properties. The exponent definitions make the $\sin(a\pm b)$ and $\cos (a\pm b)$ derivations especially nice. – Alec Dec 15 '13 at 04:14
  • @Alec $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$, I believe you are missing a factor of i. – Jeff Faraci Dec 15 '13 at 04:35
  • Haha, yes. You are right. – Alec Dec 15 '13 at 04:45
  • @boywholived There is precisely one function satisfying Alec's definition. This is a consequence of the uniqueness of the solutions to initial value problems of ordinary differential equations. – Potato Dec 15 '13 at 08:19
  • @potao . Yes, sorry. – hrkrshnn Dec 15 '13 at 08:22
  • without using l'hopital's rule, if you have $y'(0)=0$, then by definition $\lim\limits_{x\to 0} \frac{sin(x)-sin(0)}{x-0}=y'(0)=1$ – imj Dec 15 '13 at 08:25
  • I don't like this proof either. Actually it isn't a proof you just defined $\lim x\to 0 \frac{\sin x}{x}$ to be $1$ by setting $y'(0)=1, y(0)=0$ so there is nothing left to say after that. Also the use of Hopital's rule doesn't even make sense, it's circular, that thing in the limit is the derivative of $\sin x$ at $0$ by definition. – JLA Dec 15 '13 at 08:26
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    @JLA I can see why l'Hopital's is unnecessary, but how is this circular? This is totally a valid way of defining sine! – Alec Dec 15 '13 at 08:28
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    Well using Hopital's rule to show the limit is $1$ is circular, because if you don't know what $\lim_{x\to 0} \frac{\sin x}{x}$ is then you can't use Hopital's rule since you don't know what the derivative is. But yes even without this step what you did shows that $\lim_{x\to 0} \frac{\sin x}{x}=1$, but only because you essentially defined $\lim _{x\to 0} \frac{\sin x}{x}$ to be $1$, so I don't think this should be called a "proof". If this is how you define $\sin x$ then there is no reason to ask why $\lim _{x\to 0} \frac{\sin x}{x}=1$. – JLA Dec 15 '13 at 08:34
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I wonder why the geometric proof based on arc-length (or area) is termed as non-rigorous. The concept of integration itself arose from the need to rigorously define/extend the notion of length and area to figures which were not composed of straight lines. To me the usual geometric definition of $\cos t, \sin t$ as the coordinates of a point $P$ lying on the circle $x^{2} + y^{2} = 1$ with center $O$ is the simplest one and this is the route through which most beginners learn the properties of $\sin $ and $\cos $ functions. If $A = (1, 0)$ then the number $t$ (in $\cos t, \sin t$ in above definition) represents either the arc length $AP$ or twice the area of sector $AOP$. I really doubt if there is any student whose first interaction with the symbols $\sin, \cos $ starts with $$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$$

It is only in later mathematical education that one encounters the definition of $\sin x$ as an infinite series or as an infinite product or as a solution to a differential equation (and perhaps many more definitions). Note that the analytical rigorous definition of trigonometric functions which is equivalent to the above geometric version is the equation $\tan^{-1}x = \int_{0}^{x}dt/(1 + t^{2})$ for all real $x$. This is similar to defining $\log x $ as $\int_{1}^{x}dt/t$

Under the geometric definition it is very easy (especially using the concept of area rather than arc-length) to show that for $x \in (0, \pi/2)$ we have $\sin x < x < \tan x$ and then we have two results coming from it (by the way, using the geometric definition the symbol $\pi$ is defined to be the area enclosed by circle $x^{2} + y^{2} = 1$):

1) $\lim_{x \to 0}\cos x = 1$

Since $\cos x$ is even function of $x$ it is sufficient to consider $x \to 0+$ and then $1 > \cos x = 1 - 2\sin^{2}(x/2) > 1 - 2(x/2)^{2} = 1 - x^{2}/2$ so that by squeeze theorem we have $\lim_{x \to 0^{+}}\cos x = 1$.

2) $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$

Clearly from $\sin x < x < \tan x$ we get $\cos x < \dfrac{\sin x}{x} < 1$ and then by squeeze theorem $\lim_{x \to 0^{+}}\dfrac{\sin x}{x} = 1$. Since $(\sin x )/x$ is even function of $x$ considering $x \to 0^{+}$ is sufficient.

Paramanand Singh
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  • Actually I was taught the series first and then understood the circle after. This was mostly due to me self teaching calculus before learning trigonometry, so it isn't out of the question. – Sidharth Ghoshal Dec 15 '13 at 08:21
  • @frogeyedpeas: I fully accept your point. But you also need to consider that studying calculus before basic trig functions is sort of an unusual approach. For example in your approach one to has to learn limit of sequences, series, trig functions, limit of functions of a real variable, continuity and then the fact that derivative of $\sin x$ is $\cos x$. – Paramanand Singh Dec 15 '13 at 08:26
  • @frogeyedpeas: Moreover most introductory calculus texts would prefer to treat limits of functions, continuity, derivative before dealing with sequences and series. But there are books which do the other way and I believe teaching sequence & series before continuity has its merits – Paramanand Singh Dec 15 '13 at 08:28
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Use L'Hopitals Rule. The limit is of the form

$$ \lim_{x\to 0} \frac{\sin x}{x}=\frac{0}{0}. $$ Thus we can now apply L'Hopitals rule which consists of taking the derivative of the numerator and denominator and than taking the limit, thus we obtain $$ \lim_{x\to 0} \frac{\cos x}{1}=1. $$ If you are looking for something more rigorous, let me know.

Jeff Faraci
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    Given that the limit is precisely the definition of the derivative of $\sin x$ at $0$, I don't think L'H is a particularly rigorous approach here. – L. F. Dec 15 '13 at 03:40
  • @L.F. I apologize sir. If you would like a more detailed proof perhaps you can provide me with a non-relativistic definition of rigorous. Thank you for advice my friend.- Jeff – Jeff Faraci Dec 15 '13 at 03:42
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    Jeff, forget about rigor. I want a proof that is non-circular. – Adam Dec 15 '13 at 03:44
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    @Jeff No need to apologise - I am merely pointing out that the reasoning you use is a tad circular: to use L'H, we need to differentiate $\sin$ at $0$, and to do that we need the limit we are trying to find. To provide a useful answer though, the OP should say how he is defining $\sin$. – L. F. Dec 15 '13 at 03:45
  • Do you prefer a $\delta$, $\epsilon$ type proof? – Jeff Faraci Dec 15 '13 at 03:47
  • This proof is perfectly rigorous and non-circular if you are careful about your definition of $\sin$. Luckily, the OP doesn't care how we define it. – Tyler Dec 15 '13 at 04:16
  • @Tyler Yes, well said. I agree. Thanks – Jeff Faraci Dec 15 '13 at 04:17
  • I am curious. What is the definition of sine under which this proof is rigorous? – Adam Dec 15 '13 at 04:48
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    @Jeff This proof is not only not rigorous it doesn't even make sense. Under no definition of $\sin x$ can you use Hopital's rule to calculate the limit of something when the thing you are calculating is the derivative in the first place. – JLA Dec 15 '13 at 08:29
  • @JLA This method was developed by Bernoulli/L'Hopital. I am sorry you do not like it. I take it you do not enjoy Churchill's complex variables either, or mathematical operations when he discusses removable singularities. – Jeff Faraci Dec 15 '13 at 08:33
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    @Jeff No I am sorry that you do not have an understanding of what Hopital's rule is. To use Hopital's rule you need to know $\frac{d}{dx} \sin x$ at $x=0$, ie. you need to know $\lim _{x\to 0} \frac{\sin x-\sin 0}{x-0}$, ie. you need to know $\lim _{x\to 0} \frac{\sin x}{x}$, which is what you are looking for in the first place. So Hopital's rule is useless here and this proof is wrong. – JLA Dec 15 '13 at 08:38
  • The reasoning is circular. To calculate the limit you applied L'Hopital's rule, to apply L'Hopital's rule you need to know the derivative of $\sin$ at $0$, the derivative of $\sin$ at $0$ is precisely the limit you're trying to calculate. How don't you see this is a circular reasoning? – jjagmath Feb 06 '25 at 10:45