Why we assign $1/2[f(\epsilon+0) + f(\epsilon-0)]$ as the value at the discontinuity point?

My question is why do we define $f(\xi+0)$ as $\lim\limits_{\varepsilon \rightarrow 0} f(\xi + \varepsilon^2)$ and $f(\xi-0)$ as $\lim\limits_{\varepsilon \rightarrow 0} f(\xi - \varepsilon^2)$.

Thanks.

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The mystery symbol there is actually called \xi. – Qiaochu Yuan Aug 26 '10 at 5:58
We generally don't define $f(\xi+0)$ in this way: I've never seen this cheesy trick of defining one-sided limits using squares before. What is this book? – Robin Chapman Aug 26 '10 at 6:18
@Chapman: It is from Introduction to Calculus and Analysis by R.Courant and F.John. – Jichao Aug 26 '10 at 8:02
Concerning the title question, here is a quote from page 49 of Riemann's Zeta Function by H. Edwards: "In other words $\psi (x)=\sum_{p^{n}<x}\log p$ except when $x$ is a power; at the jumps $x=p^{n}$ the value of $\psi$\ is defined *as usual*, to be halfway between the new and old values $\psi (x)=\frac{1}{2}[\psi(x-\epsilon )+\psi (x+\epsilon )]$". Concerning your last question, I have never seen those definions in this way. Since $\epsilon ^2$ tends to zero fast than $\epsilon$ as $\epsilon$ approches zero, I guess that there is some connection with the derivative of function. – Américo Tavares Aug 26 '10 at 9:42
@Robin Chapman:But if we do not use squares then $\epsilon$ could be negative. – Jichao Aug 27 '10 at 12:52

Yes, or even better if $f$ is not given as such then we can redefine $f$ to have the mean value at the point of discontinuity and then $S[f]$ (the Fourier series of $f$) will converge to $f$ at that point. – AD. Aug 26 '10 at 6:00