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$$ \text{Let } \left|\sin x - 0\right| < \epsilon. \\ -\epsilon < \sin x < \epsilon \\ \arcsin (-\epsilon) < x < \arcsin (\epsilon) \\ -\arcsin \epsilon < x < \arcsin \epsilon \\ \left|x\right| < \arcsin \epsilon \\ \left|x - 0\right| < \arcsin \epsilon \\ \text{Let } \delta = \arcsin \epsilon. \\ 0 < \left|x - 0\right| < \delta \implies \left|\sin x - 0\right| < \epsilon \\ \lim_{x->0} \sin x = 0 \\ \lim_{x->0} \sin x = \sin 0 \\ \sin x \text{ is continuous at the origin} $$

In particular, is it safe to get from $-\epsilon < \sin x < \epsilon$ to $\arcsin (-\epsilon) < x < \arcsin (\epsilon)$ by applying the inverse function to all sides of the inequality? Can this operation be dangerous for some functions, functions whose inverses don't share a strictly positive or negative relation?

mcandre
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  • What definition of $\sin x$ are you using? The McLaurin series? – Ashvin Swaminathan Jul 25 '16 at 16:56
  • I'm a beginner, so I wasn't using any particular Tayor series, just trying to work from basic geometric principles. – mcandre Jul 25 '16 at 16:59
  • An argument with basic geometric principles is that for $x > 0$, you can view $\sin x$ as opposite over hypotenuse on a right-triangle, and as the angle $x$ decreases, the ratio must also decrease. Then for $x < 0$, you need to assume that $\sin x$ is an odd function. – Ashvin Swaminathan Jul 25 '16 at 17:04

3 Answers3

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Provided you know properties of the arcsine your idea will be a proof. However, are you sure you do not need to know that $\sin$ is continuous to deduce properties of the arcsine?

Spivak's calculus book has a note about a faulty proof he had in there in one of the pre-publication drafts. It used the square-root function in a proof that $x^2$ is continuous. But then, later, he used continuity of $x^2$ in the proof that the square-root exists. Fortunately, he caught the mistake before publication.

GEdgar
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Define $\sin x$ on a small neighborhood $(-\delta, \delta) \subset \mathbb{R}$ by its McLaurin series $\sin x = \sum_{n = 0}^\infty (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$. For $x \in \mathbb{R}$ with $|x|$ sufficiently small, the value of the series can be made arbitrarily small...

Ashvin Swaminathan
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You can use inequality of:

$$0\le \left| \sin { \left( x \right) } \right| \le \left| x \right| $$

then appling "squeeze theorem" directly

haqnatural
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  • How do you prove this inequality? I don't see immediately how to do it without a McLaurin series... – Ashvin Swaminathan Jul 25 '16 at 17:01
  • Thanks! I forgot about the squeeze theorem method – mcandre Jul 25 '16 at 17:03
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    @AshvinSwaminathan if $0 < x < \pi/2$ then $x/2$ is the area of a circular sector with central angle $x$ and $\sin x/2$ is the area of a triangle contained entirely within that sector. If $x > \pi/2$ the inequality is obvious. The inequality can be extended to negative $x$ without hassle. – Umberto P. Jul 25 '16 at 17:06
  • @UmbertoP. Thanks! That's really great! – Ashvin Swaminathan Jul 25 '16 at 17:07
  • @AshvinSwaminathan,look at here there're lots of applications of this inequality http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – haqnatural Jul 25 '16 at 17:12