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It is well known that

$$\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$$

I know several proofs of this: the geometric proof shows that $\cos(\theta)\leq\frac {\sin(\theta)}{\theta}\leq1$ and using the Squeeze Theorem I conclude that $\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$, other proof uses the Maclaurin series of $\sin(x)$. My question is: is there a demonstration of this limit using the epsilon-delta definition of limit?

Amr
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dwarandae
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6 Answers6

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Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|<1-\cos\theta.$$ Since $1-\cos\theta=2\sin^2\frac{\theta}{2}\le\frac{\theta^2}{2}$, hence $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|\le\frac{\theta^2}{2}.$$ Now it is easy to use $\varepsilon-\delta$ definition to get the answer.

xpaul
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For every $x \ne 0$ we have $$ \Big|1-\frac{\sin x}{x}\Big|=\Big|1-\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}\Big|\le \sum_{k=1}^\infty\frac{|x|^{2k}}{(2k+1)!}\le\frac13\sum_{k=1}^\infty\frac{|x|^{2k}}{(2k)!}=\frac{\cosh|x|-1}{3}. $$ Given $\varepsilon>0$, let $\delta=\cosh^{-1}(1+3\varepsilon)$. Then $$ 0<|x|\le\delta \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{\cosh|x|-1}{3}\le \varepsilon. $$ Another approach is to notice that $$ x-\sin x\le \frac{x^2}{2} \quad \forall\ x \in [0,\pi]. $$ Since $\sin$ is odd we have $$ -x+\sin x\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,0]. $$ Hence $$ |x-\sin x|\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,\pi]. $$ Given $\varepsilon>0$, we have $$ 0<|x|\le 2\varepsilon \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{|x|}{2} \le \varepsilon. $$

HorizonsMaths
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    It seems to me that there is a big problem with using the Taylor series. Notice that

    $$\frac{d}{dx} \sin x := \lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h} \equiv \lim_{h \to 0} \left[ \left(\frac{\cos h -1}{h}\right) \sin x+ \left(\frac{\sin h}{h}\right) \cos x \right].$$ By using the Taylor series, you are using the fact that the derivative of $\sin x$ is $\cos x$, and so are tacitly assuming that $(\sin h)/h \to 1$ as $h \to 0$.

     – Fly by Night Jan 23 '13 at 21:06
  • @Fly by Night One way to define sine is to define it by its Taylor series. (This makes the problem easier though). – Amr Jan 23 '13 at 21:33
  • @Amr Indeed, in that case the proof of the limit is trivial (it becomes an application of the laws of indices). Moreover, the need to know the limit of $\frac{1}{x}\sin x$ also disappears because we can calculate $\frac{d}{dx}\sin x$ by differentiating monomials. The "definition" of $\sin x$ as a series comes from Taylor series and comes from assuming that $\frac{1}{x}\sin x \to 1$ as $x \to 0$ and then pretending to forget where we got the series from. A little dishonest in my opinion. – Fly by Night Jan 23 '13 at 21:37
  • @Fly by Night I agree. I will try to put a solution (Its long though) – Amr Jan 23 '13 at 21:51
  • @FlybyNight One does not have to assume that [$\sin(x)/x \rightarrow 1$ as $x \rightarrow 0$] to define sine. There is no need for quotatiom marks. – user123454321 Jan 23 '13 at 22:15
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    @FlybyNight: there is more than one way to define $\sin(x)$. One is by the geometric ratio. Another is by its power series (which is being misnamed Taylor Series here). If defined as a power series, $\frac{\sin(x)}{x}$ is just a power series whose constant term is $1$ and the result is trivial. We don't need to worry about evaluating $$ \lim_{x\to0}\frac{\sin(x+h)-\sin(x)}{h} $$ using any trigonometric formulas. – robjohn Jan 24 '13 at 00:42
  • @robjohn Sorry for misnaming it. That's the term I learned in my first calculus course which was not in English! – user123454321 Jan 24 '13 at 05:41
  • @GilYoungCheong: It's not really wrong; Taylor series is a power series. However, Taylor series are derived by taking derivatives of a given function. That is, even though they are the same thing, the Taylor series is derived from the function by taking derivatives, and a power series is often used to define a function. It is a semantic difference, and perhaps the reason people resisted the use of Taylor series. – robjohn Jan 24 '13 at 07:29
  • @robjohn I realise that some people choose to define sine in terms of a power series. But that seems rather absurd to me. Where did the motivation for the coefficients of the power series come? It came from finding the Taylor series and then saying "Let's forget that we used the Taylor series." If you can show can motivate and justify the choice of power series coefficients, without simply checking that it has the same properties of the geometric definition, then I'll be won over. You could say that it's just a definition. In that case, I could just define a power series for... (cont...) – Fly by Night Jan 24 '13 at 16:28
  • @robjohn (...cont) Fly$(x) := x + x^2/2! + \cdots$, where the tail are some terms that give a convergent series. Then why aren't people on here asking about the limit of Fly$(x)/x$? It's because my series has no motivation, and no justification. The only reason people are interested in $\sin(x)/x$ is to prove that $\frac{d}{dx}\sin x \equiv \cos x$. So they must have the geometrical idea of sine in mind. It really is a circular argument. Define sine as a power series and this whole question would never have been posted. – Fly by Night Jan 24 '13 at 16:32
  • @FlybyNight, Sorry for commenting on an old post and coming out of nowhere, but since there is a geometric proof for $\sin'(x) = \cos(x)$ that doesn't involve the limit of $\sin(x)/ x$, assuming that proof, you can use the derivative and hence the Taylor series to prove this limit without dealing with a circular argument. – Stefan Octavian Jun 25 '19 at 07:44
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Taking as the definition per the OP's comment, $$\sin x = \sum_{n=0}^\infty {(-1)^n x^{2n+1} \over (2n+1)!}$$ we have that $${\sin x \over x} = \sum_{n=0}^\infty {(-1)^n x^{2n} \over (2n+1)!} =1 - {x^2 \over 6} + {x^4 \over 120} - \cdots \,,$$ which is an alternating series for all $x$ and with decreasing terms for, say, $|x| < 1$.

Ok, so let $\epsilon>0$ be arbitrary. Let $\delta = {\rm min}\{\sqrt{6\epsilon},1\}$. Assume $|x| < \delta$. Then $$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}$$ since the series is alternating with decreasing terms, and therefore $$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}< {\big(\sqrt{6\epsilon}\big)^2 \over 6}=\epsilon$$ Thus the limit is $1$.

Michael E2
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Define $\sin(x) := x - x^{3}/3! + x^{5}/5! - \cdots$ and show that this is an analytic function and see that we can take derivative term by term so that $\sin'(x) = 1 + x \cdot f(x)$ for some continuous $f$. We have the required limit $= \sin'(0) = 1$. This does not involve any geometric argument and you can trace all the process until you meet $\epsilon$ and $\delta$.

ALSO: See Walter Rudin's Principles of Mathematical Analysis Theorem 8.1.

user123454321
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  • The definition of $\sin x$ needs some justification. – Fly by Night Jan 23 '13 at 21:08
  • @FlybyNight I didn't say that I included all the details. For me, OP's question seemed to ask how one would prove that identity more rigourously, using analysis. – user123454321 Jan 23 '13 at 21:17
  • The OP's question specifically states that he knows of "[an]other proof [that] uses the Maclaurin series of $\sin(x)$" and that his question was "...is there a demonstration of this limit using the epsilon-delta definition of limit?" – Fly by Night Jan 23 '13 at 21:19
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    @FlybyNight And since it is easy to see that the sum is convergent for all real $x$, I don't see the necessity for "justification of definition," because it is well-defined. Are you talking about relating more intuitive and geometric definition to this one? Even OP commented that he/she is using summation definition. What is there to justify? Only thing that you should justify is whether you can take term by term derivative, which you can find in the reference. – user123454321 Jan 23 '13 at 21:22
  • @FlybyNight Good point, I missed that phrase when I was reading the post. Thanks for pointing it out. However, each step for the proof that I made is indeed from $\epsilon, \delta$ definition, if you understand the summation rigorously. Therefore, I think it is essentially a proof, using $\epsilon, \delta$ definition. – user123454321 Jan 23 '13 at 21:25
  • It needs justification because it comes from knowledge of the derivative. But, as mention by others, using the derivative is questionable because you can't show that $\frac{d}{dx}\sin x \equiv \cos x$ without showing that $(\sin x)/x \to 1$ as $x \to 0$. In short: any "proof" involving Taylor series raises more questions than it answers. – Fly by Night Jan 23 '13 at 21:25
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    @FlybyNight Note that I did not use cosine. In the proof, we have $f(x) = -(x^{2}/3! - x^{4}/5 + \cdots)$. There is no need for quotation marks. – user123454321 Jan 23 '13 at 21:26
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    A definition does not need justification. We can define $f(x)=\sum_{i=1}^{\infty}(-1)^i x^{(2i+1)}/(i+1)!$ at any point. The justification comes when we want to say that $f(x)=\sin (x)$ when someone else has already picked a definition for $\sin(x)$. But if there is no preexisting definition then there can be no contention as to how I choose to define $\sin(x)$. I think there are problems with definition the function $\sin(x)$ from geometry, because you're already leaning on equivalence of angles mod $2 \pi$ to able to extend the domain to all real numbers. – R R Mar 15 '14 at 22:53
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    Then there is the question of extending the domain to all complex numbers from the geometric definition which I have never seen done. $\sin(x)$ exists as an analytical animal first and foremost when we treat it as a function. – R R Mar 15 '14 at 22:54
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I believe you are looking for a a proof without differentiation but only metric spaces.

Define $e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$, define $sin(z) = \frac{1}{2i}(e^{iz}-e^{-iz})$

$\frac{\sin(z)}{z} = 1 +\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}$

Now we prove the sum of the latter term goes to 0:

$0\le|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{6^{2k}}|=\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}$

As $z\rightarrow 0$, we can pick $\displaystyle N\in\mathbb{N}.\;\forall n\ge N.\;|z_n|<6\;\Longrightarrow |\frac{z_n}{6}|<1$

$\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}} \sum_{k =1 }^\infty |\frac{z}{6}|^{2k} = \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}(\frac{1}{1-|\frac{z}{6}|}-1) = 1-1 =0$

Here we are comparing it with geometric sum. When its absolute value is sandwiched between $0$, the term has to go to $0$.

$\Longrightarrow$ $0\le\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$$|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|$ $\le$ $\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$ $\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}=0$

$\Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 0$

$\Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\frac{\sin(z)}{z} = 1 +\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 1$

mez
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  • The definition of $e^z$ needs some justification. Also, doesn't the proof that $e^{iz} = \cos z + i \sin z$ rely on Taylor series and therefore involve differentiation? – Fly by Night Jan 23 '13 at 21:09
  • It depends on how you build your theory. $e^z$ needs to be proven uniform convergent in $\mathbb{C}$ to begin with, then if you define $cos$ and $sin$ with $exp$, then you don't need to prove Euler's formula, because you used it in definition. What need proven is existence of $\pi$, periodicity and such so that this $cos$ and $sin$ defined matches the geometric interpretation of what we learnt in high school, which is troublesome. – mez Jan 23 '13 at 21:22
  • But then surely, if we define $\sin z = \frac{1}{2i}(e^{iz}-e^{-iz})$ then the fact that $(\sin z)/z \to 1$ as $z \to 0$ is trivial, and moreover, the need to know the limit of $(\sin z)/z$ disappears because we can calculate $\frac{d}{dz} \sin z$ from its series definition. – Fly by Night Jan 23 '13 at 21:32
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    @FlybyNight: actually, you can prove that $\cos(x)+i\sin(x)=e^{ix}=\lim\limits_{n\to\infty}\left(1+\frac{ix}{n}\right)^n$ as shown here. – robjohn Jan 23 '13 at 22:14
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    @robjohn In equation [7] you use the fact that $(\tan x)/x \to 1$ as $x \to 0$. This is equivalent to proving that $(\sin x)/x \to 1$ as $x \to 0$. We're back to the same old problem. – Fly by Night Jan 24 '13 at 16:43
  • @robjohn I have to agree with FlybyNight here, if there is nice geometric argument of $\lim_{x\to 0}\frac{\tan{x}}{x} = 1$, then we can use nice geometric argument to prove the problem to begin with. btw how did you type the lim with x\to 0 in the bottom while not using environment? did you use \displaystyle? – mez Jan 24 '13 at 17:30
  • @mezhang: sorry, I deleted my previous comment because it was in response to a deleted comment, so it was obsolete. For the limit I used \lim\limits_{x\to0}. If you right click on a piece of MathJax, there should be a contextual menu that lets you see the TeX Commands. – robjohn Jan 24 '13 at 17:41
  • @robjohn thanks a lot for the tip – mez Jan 24 '13 at 17:52
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$\forall \epsilon > 0$ $ \exists \delta > 0$ such that $|\theta| < \delta \Rightarrow |\frac {\sin(\theta)}{\theta} - 1| < \epsilon$

$\cos(\theta)\leq\frac {\sin(\theta)}{\theta}\leq1$

$0\leq 1-\frac {\sin(\theta)}{\theta} \leq 1-\cos(\theta) < \epsilon$

$\cos(\theta) > 1 - \epsilon$

When $\delta < \arccos (1-\epsilon)$, $|\theta| < \delta \Rightarrow |\frac {\sin(\theta)}{\theta} - 1| < \epsilon$

Doug M
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