Question

This question is inspired from this one: http://openstudy.com/study?login#/updates/4fa7f4e2e4b059b524f3bb68

How could one prove that \(\sin\) is continuous, using \(\varepsilon\), \(\delta\) arguments?

Answer 1

This is what we are trying to prove:

\[\forall\varepsilon>0,\exists\delta>0\ni|x-c|<\delta\implies|\sin x -\sin c|<\varepsilon,\]for some \(c\in\mathbb{R}\).

Answer 2

I just had an idea, but I am not sure if it is rigorous enough:

From the Mean Value Theorem, we have that\[\cos c=\frac{\sin b-\sin a}{b-a}\implies(b-a)\cos c=\sin b-\sin a\implies\sin b-\sin a\leqslant(b-a),\]since \(|\cos|\leqslant1\). Can I use this fact to say that\[|\sin b-\sin a|\leqslant|b-a|<\delta=\varepsilon,\]for every \(\delta,\varepsilon>0\)?

Answer 3

seems all right ... and better than what i've thought!!