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?
 A: 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
$$
0<|x|\le 2\varepsilon \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{|x|}{2} \le \varepsilon.
$$
A: 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 $1$.
A: 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.
A: 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\}}$$|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|$ $\le$ $\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$ $\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}=0$
$\Longrightarrow
 \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 0$
$\Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\frac{\sin(z)}{z} = 1 +\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 1$
A: Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get
$$\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.
A: $\forall \epsilon > 0$ $ \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 $\delta < \arccos (1-\epsilon)$, $|\theta| < \delta \Rightarrow |\frac {\sin(\theta)}{\theta} - 1| < \epsilon$
