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$\lim_{n\to \infty}\sin^n(x) = 0 \:\:\:\: \forall x\:\in\:[0,\pi/2) \\$

Since the pointwise limit is continuous $\forall x\:\in\:[0,\pi/2) $ therefore by definition of uniform convergence we have : $$\lim_{n\to \infty}\{\sup_{0\leq x<\pi/2} |f_n(x)-f(x)|\} = \lim_{n\to \infty}\{\sup_{0\leq x<\pi/2}|f_n(x)-0|\}\:\: \:= \lim_{n\to \infty}\{\sup_{0\leq x < \pi/2} |\sin^n(x)|\}\:\: \:=\:\lim_{n\to \infty}1\:\:=\:\:1\\\neq0$$

Thus the function should not be uniformly convergent.But it is uniformly convergent on the given interval. That means the supremum must be $0$ instead of $1.$ But i don't see why that should be the case. Can somebody please help and clarify the doubt.

Michael Hardy
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under_root
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  • Why does your function uniformly convergent on given interval? Is there another reason justifying your claim? – Hanul Jeon Apr 19 '18 at 04:36
  • using the epsilon definition, we can always choose a bigger enough n, for which |f_n(x)-0| < e , since 0<=sin(x)<1 . sorry i am new here. learning latex. – under_root Apr 19 '18 at 04:39
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    The sequence is not uniform convergent, you just showed that. –  Apr 19 '18 at 04:40
  • i was too thinking the same thing, but i read somewhere it is uniformly convergent and have been confused since then. finally posted here...thank you for clarifying... – under_root Apr 19 '18 at 04:41
  • Please define "somewhere". – Did Apr 19 '18 at 06:33
  • https://math.stackexchange.com/questions/268235/uniform-convergence-of-few-sequence-of-functions?rq=1 – under_root Apr 19 '18 at 18:04

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