# Thread: prove sin(x) is uniformly continuous.

1. Originally Posted by redsoxfan325
That is true, yes, but I think you need to show that $\displaystyle |\sin x-\sin y|\leq|x-y|$, not just state it.
Reasoning geometrically...

2. no, just use the MVT and that's all.

3. Originally Posted by Krizalid
no, just use the MVT and that's all.
Not allowed to use it though.

Showing this is not a problem though:

$\displaystyle |\sin(x) - \sin(y)| \leq |x - y|.$

There is an argument in my book that uses a unit circle to get this estimate, so that part is pretty much given, or were you reffering to my $\displaystyle \epsilon - \delta$ argument?

4. i'm sorry, but are you taking analysis?

i'm just using a fact taken from calculus I, if you can't use it, then this definitely makes no sense to me.

5. In many introductory analysis courses, students are not allowed to use facts that they know are true (like from previous Calculus courses) until they have been proven rigorously either in class or as an exercise in the homework.

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# uniformly continuous functions examples sinx

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