calculus - 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?

Answer

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.