# prove sin(x) is uniformly continuous.

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• Oct 25th 2009, 10:33 PM
cgiulz
Quote:

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...
• Oct 26th 2009, 04:40 AM
Krizalid
no, just use the MVT and that's all.
• Oct 26th 2009, 06:46 AM
cgiulz
Quote:

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?
• Oct 26th 2009, 08:10 AM
Krizalid
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.
• Oct 26th 2009, 09:57 AM
redsoxfan325
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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