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This is a simple question. Given a real valued rational function $$ f (x) = \frac{p(x)}{q(x)}\quad x\in\mathbf R^N, $$ this is called regular on a point if the denominator $q$ does not vanish there. We know that

  1. $f$ is regular on an open convex set $D$ and
  2. for a point $x_0$ on the boundary $\partial D$ of $D$ the limits of $f$ along any ray coming from $\partial D$ upon $x_0$ exists and do not depend on the ray. More precisely: for any $x_1$ from $D$ the limit $tx_1 + (1 - t)x_0$ as $t$ tends to $0$ exists and is independent of $x_1$.

Question: is $f$ regular at $x_0$? Thanks in advance.

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  • $\begingroup$ If the denominator vanishes, the function is not just irregular, it is undef ined. $\endgroup$
    – Michael Greinecker
    Commented Jul 5, 2023 at 8:05
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    $\begingroup$ @MichaelGreinecker ... Irregular points could possibly have $p(x_0)=q(x_0)=0$, so Vestic asks if this could happen under conditions (i) and (ii). $\endgroup$
    – Gerald Edgar
    Commented Jul 5, 2023 at 8:20
  • $\begingroup$ @GeraldEdgar technically under the definition given, if $q(x_0) = 0$ then $f$ is not regular at $x_0$. $\endgroup$
    – user44191
    Commented Jul 5, 2023 at 10:55
  • $\begingroup$ @user44191 : good point. It should be replaced by "can be made regular". $\endgroup$
    – Mikhail Katz
    Commented Jul 5, 2023 at 12:35

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If $f(x,y)=\frac{y-x^2}{y+x^2}$ and $D$ is any set above the parabola $y=x^2$, then it seems to me that the limit along any ray will always be $1$.

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  • $\begingroup$ Thanks to a fine example, which indicates how to strengthen the conditions to obtain a positive answer. Indeed, $\endgroup$
    – Veselic
    Commented Jul 5, 2023 at 11:20
  • $\begingroup$ Thanks very much to professor Katz for the illuminating example. $\endgroup$
    – Veselic
    Commented Jul 5, 2023 at 12:33
  • $\begingroup$ sorry for my first comment (mistake). In fact there is no possibility to extend by mere continuity either.. $\endgroup$
    – Veselic
    Commented Jul 5, 2023 at 12:36

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