Talk:Pointwise convergence

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I'm not quited convinced that the series {$$x^n$$} is not uniformly convergent on the interval [0,1).The limit function f(x) is not continious at x0=1. For every 0inf) [sup {|fn(x)-f(x)|} ]=0


 * No, that last assertion is false. The value of
 * $$ \sup \{\,|f_n(x) - f(x)| : 0 \le x \le 1 \,\} $$
 * is 1, for every n. That is because, for any fixed value of n, xn can be made as close to 1 as desired by making x close enough, but not equal, to 1.  So fn(x) approximates 1 while the limiting function f(x) remains 0, as x approaches 1. Michael Hardy (talk) 15:39, 5 March 2008 (UTC)
 * is 1, for every n. That is because, for any fixed value of n, xn can be made as close to 1 as desired by making x close enough, but not equal, to 1.  So fn(x) approximates 1 while the limiting function f(x) remains 0, as x approaches 1. Michael Hardy (talk) 15:39, 5 March 2008 (UTC)
 * is 1, for every n. That is because, for any fixed value of n, xn can be made as close to 1 as desired by making x close enough, but not equal, to 1.  So fn(x) approximates 1 while the limiting function f(x) remains 0, as x approaches 1. Michael Hardy (talk) 15:39, 5 March 2008 (UTC)


 * whoa, I am pretty sure that is not true. For any value x < 1, as n goes to infinty, x^n goes to 0. So that function does converge uniformly. —Preceding unsigned comment added by 126.109.110.248 (talk) 12:15, 13 March 2011 (UTC)