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Using the definition of removable discontinuity from Wikipedia, why can't a monotone function have this type of discontinuity?

In other words, if $x_0\in D(f)$ is a point where the monotone function $f$ is discontinuous, and if $\lim_{x\to x_{0^-}}f(x)=L^-$ and $\lim_{x\to x_{0^+}}f(x)=L^+$ why cannot be $L^+=L^-$

I've been baffled by this for far too long now, thanks for any help!

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    Only if $f(x_0)=L^+=L^-$.2012-12-29

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An increasing function can't have a removable discontinuity at points in its domain. Indeed observe that for $\epsilon>0$, $ \exists \delta>0:0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon\iff L-\epsilon For $a-\delta, $f(x) and so $ L-\epsilon Similarly for $a, $f(x)>f(a)$ and $f(a) Therefore, $\left|f(a)-L\right|<\epsilon$ for arbitrary $\epsilon>0$ and so $f(a)=L$

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    Thanks, now I understand it completely.2012-12-29
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Take $f$ and increasing function.

Let $\lim\limits_{x\to x_0^-}f(x) = a$

$f(x_0) = b$

$\lim\limits_{x\to x_0^+}f(x) = c$

Since $f$ is not continuous at $x_0$, you have $a\not=b$ or $b\not=c$. Since it is increasing, you have $a\le b \le c$ so $a or $b. And you can easily conclude that $a so $a \not= c$, ie $\lim\limits_{x\to x_0^-}f(x) \not= \lim\limits_{x\to x_0^+}f(x)$

And for a decreasing function, you just use that property for $-f$.

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    @CalvinLin Have I defined $x_0$ to be a point of *removable* discontinuity? I'm not aware of that! Anyway, this answer helped me understand the general gist of the idea, so thanks for it nontheless.2012-12-29