How can we prove that $\frac{\sin x}{x}$ is uniformly continuous at open interval $(0, 1)$ without using the mean value theorem?
I've seen a lot of answers using MVT, but I cannot find how to do it without MVT
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https://math.stackexchange.com/a/259767/42969, https://math.stackexchange.com/q/2119595/42969, https://math.stackexchange.com/q/436825/42969 – Martin R Jun 16 '24 at 09:01
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I think the question asks for a bare hands proof, not using derivatives or integrals. The second link provided by @Martin R appears (I did not check all the details) to have such a proof. – Steen82 Jun 16 '24 at 11:49