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Show that the sequence of functions $f_n(x) = x/n$ converges pointwise but not uniformly on $[0,\infty)$

I am not sure if my method for the proof is correct. Also for showing that $f_n(x)$ is not uniformly convergent, I took the negation: $\exists \epsilon > 0$ $\forall N$ $\exists n>N$ $\exists x \in \mathbb{D}$ $\Rightarrow |f_n(x) - f(x)| \geq \epsilon$

proof

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    $\begingroup$ Not sure what you are asking - are you looking for a counter example or were you supposed to include an actual example sequence which converges pointwise but not uniformly. Its also necessary to include the range of the functions since they may converge in one range but not another. $\endgroup$ – user247608 Apr 18 '16 at 6:59
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The pointwise-convergence is easy to show.

To show the convergence on your set is not uniform, and given that you have a sequence of continuous functions, it is enough to show that the limit function $f(x)$ is discontinuous - thus, the convergence could not have been uniform.

Can you show this?

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