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I’m looking for a proof of $ \displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $ that does not use other trigonometric functions or any first-order approximation to the sine function. Is this possible? The other proofs that I’ve seen on this website don’t really fit these stringent requirements, so I was hoping to see a different kind of demonstration altogether. Thanks!!

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The limit should read $ \displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $. – Haskell Curry Feb 23 '13 at 20:37
What definition of sin(x) are you taking here? – anonymous Feb 23 '13 at 20:39
Oh, so you don't want to use $\tan $ and $\cos$, for instance...Otherwise, this would do: – 1015 Feb 23 '13 at 20:39
@Azmoti: It’s not an exact duplicate because the OP seems to want to avoid Taylor series. :) – Haskell Curry Feb 23 '13 at 20:43
A functional approach is taken in "A New Approach to Circular Functions, II and lim (sin x)/x" by Gerson B. Robison in Mathematics Magazine, Vol. 41, No. 2 (Mar., 1968), pp. 66-70 – Byron Schmuland Feb 23 '13 at 20:51
up vote 9 down vote accepted

An interesting proof that sees no trigonometry or Maclaurin series until the very final conclusion (at the cost of doing Riemann integration):

Define a function $ \theta: [-1,1] \to \mathbb{R} $ as follows: $$ \forall a \in [-1,1]: \quad \theta(a) \stackrel{\text{def}}{=} \int_{0}^{a} \sqrt{1 - x^{2}} ~ d{x}. $$ For each $ a \in [-1,1] $, we can interpret $ \theta(a) $ as the signed arc-length of the parametrized curve $$ \left\{ \left( x,\sqrt{1 - x^{2}} \right) \in \mathbb{S}^{1} ~ \Big| ~ \text{$ x $ is between $ 0 $ and $ a $} \right\}. $$

Observe that $ \theta $ has the following properties:

  • $ \theta(0) = 0 $ and

  • $ \theta: [-1,1] \to \mathbb{R} $ is a strictly increasing continuous function.

There is thus an inverse function $ \theta^{-1}: \text{Range}(\theta) \to [-1,1] $, and we define the sine function (or at least part of it) by

$$ \sin \stackrel{\text{def}}{=} \theta^{-1}. $$

Hence, \begin{align} \lim_{a \to 0} \frac{\sin(\theta(a))}{\theta(a)} &= \lim_{a \to 0} \frac{{\theta^{-1}}(\theta(a))}{\theta(a)} \\ &= \lim_{a \to 0} \frac{a}{\theta(a)} \\ &= \lim_{a \to 0} \frac{a}{\displaystyle \int_{0}^{a} \sqrt{1 - x^{2}} ~ d{x}} \quad (\text{By definition.}) \\ &= \lim_{a \to 0} \frac{1}{\sqrt{1 - a^{2}}} \quad (\text{By l’Hôpital’s Rule.}) \\ &= 1. \end{align} Therefore, $$ \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1. $$

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Cannot l’Hôpital’s Rule be used directly on $\sin(x)/x$? – enzotib Feb 23 '13 at 20:54
@enzotib: If you use l’Hôpital’s Rule on $ \sin $, you get $ \cos $, which the OP wants to avoid. :) – Haskell Curry Feb 23 '13 at 20:55
@enzotib, We aren't allowed to use l’Hôpital’s Rule because that would assume that $\frac{d}{dx} \sin x = \cos x$. However, that result follows from $\lim_{x\to 0} \frac{\sin x}{x} = 1$, so by assuming $\sin\prime = \cos$, we are assuming the limit we want to prove. – George V. Williams Feb 23 '13 at 20:56
Not necessarily. We typically take the definition of sin(x) as a power series. From this definition, it's straightforward to prove that sin'(x) = cos(x) without proving that sin(x)/x tends to 1. – anonymous Feb 23 '13 at 21:46
@anonymous, you cannot take sine as a power series without first knowing the derivative of sine. – CogitoErgoCogitoSum Feb 24 '13 at 1:22

Taking the definition $$\sin(x)=\frac{1}{2i} (e^{ix} - e^{-ix}) $$

Than you get from the exponentialfunction the identity $$\sin(x)=\sum_{k=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp \cdots$$

So we have $$\lim_{x\rightarrow 0 } \frac{\sin(x)}{x}=\lim_{x\rightarrow 0} 1-\frac{x^2}{3!} + \frac{x^4}{5!}\mp \cdots=1$$

Else you could define $$\sin(x)=x\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2 \pi^2 } \right)$$ Than $$\lim_{x\rightarrow 0 } \frac{\sin(x)}{x} =\lim_{x\rightarrow 0} \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right)$$

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I think you could, instead of expanding the exponential function, directly take the limit of $e^{ix}/x$, with the aid of L'Hôpital. – awllower Feb 24 '13 at 3:43

If you're not defining $\sin$ via Taylor series, the obvious other rigorous alternative would be to define it (near $0$) as the inverse function of $\arcsin$, where we take $\arcsin y = \int_0^y \frac{dt}{\sqrt{1-t^2}}$. Then

\begin{eqnarray} \int_0^y \frac{dt}{\sqrt{1-t^2}} &<& \int_0^y \frac{dt}{\sqrt{1-y^2}} = \frac{y}{\sqrt{1-y^2}} \\ \int_0^y \frac{dt}{\sqrt{1-t^2}} &>& \int_0^y \, dt = y \end{eqnarray}

So $\sqrt{1-y^2}<\frac{y}{\arcsin y}<1$, and thus $\lim_{y \to 0} \frac{y}{\arcsin y} =1$ by the squeeze theorem.

After making the change of variables $x=\arcsin y$ in this last limit (justified because $\arcsin y$ is continuous and monotone), it follows that $\lim_{x \to 0} \frac{\sin x}{x}=1$.

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why it is so obvious that your function is injective ? – Dominic Michaelis Feb 23 '13 at 21:09
@DominicMichaelis: It's the integral of something positive. – Micah Feb 23 '13 at 21:10

I'm assuming that OP uses the geometric (slightly non-rigorous) definition of the trigonometric functions, i.e., $(\cos(t),\sin(t))$ is the point on the unit circle forming angle $t$ with the positive $X$ axis).

The requested limit (known as the first fundamental limit) can't be computed by any method that relies on the fact that $\sin'(x)=\cos(x)$ (in fact, this limit is required in the proof of that fact), and I think this is the point of this exercise. Here is a geometric proof, legitimate along these lines.

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You just hit the nail on the head, Ittay. However, if we can give a formal definition of ‘angle’, then the problem of semi-rigor is solved. One can define ‘angle’ using arc-length, which is what I did above, but of course, I needed to resort to Riemann integration. – Haskell Curry Feb 24 '13 at 3:54
and you just drove the nail all the way into the wood @HaskellCurry :) – Ittay Weiss Feb 24 '13 at 4:03

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