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?
Best Answer
Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|<1-\cos\theta.$$ 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.