# Prove $\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$ with the epsilon-delta definition of limit.

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?

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A proof with Maclaurin series sounds somehow debatable or at least strange - you need derivatives to write this series and the question itself is about the derivative of $\sin$ at $0$. –  savick01 Jan 23 '13 at 20:31
what definition of sin –  user58512 Jan 23 '13 at 20:32
I think the infinite sum is more rigourous definition of sine function than geometric hand waving. It is not debatable. –  GYC Jan 23 '13 at 20:32
To expand what @user58512 says, if you use a geometric definition of $sin x$, then you should expect a proof with a geometrical basis. One could modify the proof using the Squeeze Theorem to conform to the epsilon-delta definition. So again, what definition of sine? –  Michael E2 Jan 23 '13 at 20:39
@MichaelE2 the infinite sum. –  dwarandae Jan 23 '13 at 20:41

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

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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.

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The definition of $\sin x$ needs some justification. –  Fly by Night Jan 23 '13 at 21:08
@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. –  GYC Jan 23 '13 at 21:22
@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. –  GYC Jan 23 '13 at 21:26
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 at 22:53
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 at 22:54

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 $1$.

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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 $$0<|x|\le 2\varepsilon \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{|x|}{2} \le \varepsilon.$$

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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: 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 0:42

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

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

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

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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
@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
@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