Can someone please explain to me why sin(x^2) is not uniformly continuous.
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Originally Posted by summerset353 Can someone please explain to me why sin(x^2) is not uniformly continuous. On what set?
the set being R
The reason is that the function is contionuos and bounded but its derivative is unbounded... Kind regards
that's false. consider being uniformly continuous on but its derivative is not bounded.
Krizalid is right. Having bounded derivative is a sufficient condition for being uniformly continuous, but it is not necessary. Consider the sequences , . We have that tends to 0 and . From this fact it follows that is not uniformly continuous.
Originally Posted by summerset353 Can someone please explain to me why sin(x^2) is not uniformly continuous. Let . Let and . Notice that and can be arbitrarily close (for n large enough) and yet .
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