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- Thread starter ForMyThunder
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HallsofIvy

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Given any [itex]\epsilon> 0[/itex], there exist N such that if n> N, [/itex]|a_n(x)- a(x)|< \epsilon[/itex] for all x.

Well, [/itex]|(a_n(x)+ 1)- (a(x)+ 1)|= |a_n(x)- a(x)|[/itex] for all x! So it immediately follows that [itex]a_n(x)+ 1[/itex] converges uniformly to a(x)+ 1.

(Oh, and it really does not make sense to talk about a sequence of

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