Any $\varepsilon$-$\delta$ proof of this limit statement using only the trigonometric or circular definition of the sine function?

Question

If $x$ is in radians, then we know that $$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$ An elementary "proof" involving a geometric construction is often found in calculus texts.

And, one can also supply a rigorous proof of this using the machinary of uniformly convergent series of functions. Am I right?

Now I'm wondering if it is possible to give an $\varepsilon$-$\delta$ proof of this statement but using only the trigonometric or circular definition of the sine function.

If so, then can anybody in the valued Math SE community please supply such a proof in a detailed anough answer? Thanks in advance!!

Answer 1

$x \in (-\frac \pi2, \frac \pi2) \implies |\sin x| \le |x| \le |\tan x|$

$1 \le |\frac {x}{\sin x}| \le |\sec x|$ and in this interval of $x,$ $x$ and $\sin x$ have the same sign.

$1 \le \frac {x}{\sin x} \le \sec x\\ 1 \ge \frac {\sin x}{x} \ge \cos x$

$\forall \epsilon > 0, \exists \delta > 0: |x|<\delta \implies |\frac {\sin x}{x} - 1| < \epsilon$

$|\frac {\sin x}{x} - 1| \le 1-\cos x $

let $\delta = \min (\frac \pi2, \arccos (1-\epsilon))$