Sorry for the naff title, don't really know how else to put it. The question goes... Let be a sequence tending to and let be a real number. Give an proof that . Looks simple enough, but I can't get a handle on it at all.
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Originally Posted by chella182 Sorry for the naff title, don't really know how else to put it. The question goes... Let be a sequence tending to and let be a real number. Give an proof that . Looks simple enough, but I can't get a handle on it at all. Hi chella182. Given there is such that for all The result follows immediately as
In general if we have two sequence and with... , ... and is a continuos function is... In your case is ... Kind regards
Originally Posted by TheAbstractionist Hi chella182. Given there is such that for all The result follows immediately as Is that it? That the solution? :S
Originally Posted by chella182 Is that it? That the solution? :S Hi chella182. Yes. Let and Then you want that you can find some such that right? That was what I proved.
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