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My sequence of functions $$f_n (x) = \begin{cases} 1 & ,x = \frac{1}{n} \\ x & ,x = 1,1/2, ...,1/(n-1) \\ 0 & ,otherwise \end{cases}$$

My attempt is to fix $k \in \mathbb{N}$, consider the following cases when $x = 1/k$ for $n \geq k $ and $x \neq 1/k$. Is there a better approach to find the pointwise limit $f$ of this sequence $f_n$?

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    $\begingroup$ Pointwise limits must be calculated pointswise. So, yes, you have to split up in these cases. $\endgroup$ – user370967 Nov 19 '18 at 17:59
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Let $x$ be a real.

If $x=\frac 1k$ then for large enough $n\ge k+2,$ we will have

$$\frac 1n <\frac{1}{n-1}<x\le 1$$ then $$f_n(x)=x$$

and if $x\ne \frac 1k \implies f_n(x)=0$

The pointwise limit function is $$f:x\mapsto x$$ if $x=\frac 1k$ and zero elsewhere.

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