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Given closed interval $[a,b]$. How to show that if we have a sequence of continuous functions $\{f_n\}$ point wise converging to $f$, then $f$ does not have infinite discontinuities?

By infinite discontinuity, I mean the one-sided limits don’t go to $+\infty$ or $-\infty$

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Let $f_n$ being equal to $\frac 1 x$ over $(\frac 1 n,1]$ and joining linearly $0$ to $n$ between $x=0$ and $x=\frac 1 n$. In particular, $f_n$ are continuous.

That function sequence converges pointwise to $f$ being $\frac 1 x$ on $(0,1]$ and $0$ at $x=0$, which is not bounded.

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  • $\begingroup$ Looking at the last sentence "one sided-limits don't ...", this is unlikely to be what he meant. "infinite discontinuities" rather means that the discontinuity is due to the function being unbounded around some point. @MariosGretsas $\endgroup$
    – nicomezi
    Oct 29, 2019 at 15:21
  • $\begingroup$ @nicomezi yes, if the function does not have infinite discontinuities on a closed interval, then it’s bounded. This is what I want. $\endgroup$
    – user651267
    Oct 29, 2019 at 15:28
  • $\begingroup$ @nicomezi i understood infinitely many discontinuities..my bad..sorry $\endgroup$
    – Marios Gretsas
    Oct 29, 2019 at 15:29
  • $\begingroup$ I understood the same as you at first, infinite discontinuity is not a very common name. ;) @MariosGretsas $\endgroup$
    – nicomezi
    Oct 29, 2019 at 15:32
  • $\begingroup$ @nicomezi...i know ..;) $\endgroup$
    – Marios Gretsas
    Oct 29, 2019 at 15:32

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