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$f_n(x):[0,1]\rightarrow \mathbb{R}$ defined by $$f_n(x)= \sin(n\pi x)$$ if $x\in [0,1/n]$, and $$f_n(x)=0$$ if $x\in (1/n,1]$ Then

  1. It does not converge pointwise.

  2. It converges pointwise but the limit is not continous.

  3. It converges pointwise but not uniformly.

  4. It converges uniformly.

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up vote 3 down vote accepted

Here is a hint for your problem:

Note that given an $n$ you will always be able to find a $c \in [0, \frac{1}{n}]$ such that $f_n(c) = 1$.

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we can chose $c={1\over 2n}\in[0,{1\over n}]$ – Un Chien Andalou Jun 18 '13 at 5:26

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