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

Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|<1-\cos\theta.$$ But $1-\cos\theta=2\sin^2\frac{\theta}{2}\le\frac{\theta^2}{2}$ and 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.

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

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 $\delta$.

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