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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!!

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marked as duplicate by projectilemotion, Namaste calculus Feb 15 '17 at 0:37

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  • $\begingroup$ By the trigonometric definition, do you mean "opposite" / "hypotenuse"? That seems pretty geometric to me. As does the circular definition, come to think of it. $\endgroup$ – Bobbie D Feb 14 '17 at 21:17
  • $\begingroup$ Can you be more specific about exactly what definition you are referring to? $\endgroup$ – Hans Lundmark Feb 14 '17 at 21:18
  • $\begingroup$ Does any of the proofs here satisfy your criterion: math.stackexchange.com/questions/75130/…? $\endgroup$ – Hans Lundmark Feb 14 '17 at 21:19
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    $\begingroup$ The alternative proofs which are not geometric generally either rely on defining sine by its power series or as a solution to a differential equation, both of which I feel you'd consider circular reasoning (though they are not as they really just depend on how you initially define sine). $\endgroup$ – adfriedman Feb 14 '17 at 21:23
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    $\begingroup$ This is meaningless unless you define what $\sin$ is. Only then do we have a solid basis to answer your question. $\endgroup$ – MathematicsStudent1122 Feb 14 '17 at 21:32
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$x \in (-\frac \pi2, \frac \pi2) \implies |\sin x| \le |x| \le |\tan x|$

$1 \le |\frac {x}{\sin x}| \le |\sec x|$ 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 $\delta = \min (\frac \pi2, \arccos (1-\epsilon))$

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