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My teacher proceeds to tell us it must be the points in which it is not defined: $x^2+y^2-1=0$ which is a circle $[0,0], r=1$.

Why isn't it continuous there? Undefined point doesn't imply discontinuity, or does it?

gerw
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SlowerPhoton
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  • I agree on that. – Hagen von Eitzen Oct 04 '17 at 06:29
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    I am just "thinking out loud" continuity is when the limit at a point is the value of the function in a point. So does it mean that if a function is not defined in a point it implies that it have discontinuity? No has by definition there is a limit also if the point is not defined, in this case, I think that it has something to do with the fact that the denominator is defined (and its value is $0$) but the limits are ($\pm \infty$) – gbox Oct 04 '17 at 06:48
  • well, how COULD it be continuous if the point doesn't exist? – Saketh Malyala Oct 04 '17 at 06:50
  • there are such things as removable discontinuity --- but they aren't continuous -- hence "discontinuity" – Saketh Malyala Oct 04 '17 at 06:50
  • @Saketh Malyala after some reasearch it seems you are right. – SlowerPhoton Oct 04 '17 at 07:02
  • Please use appropriate tags. This is not functional analysis. – gerw Oct 04 '17 at 18:09

2 Answers2

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If you have $n(x,y)$ continuous and $d(x,y)$ continuous and $d$ does not annulate then the fraction $f(x,y)=\dfrac{n(x,y)}{d(x,y)}$ is continuous.

I have shown it here for a simplified version $\dfrac{f(x)}{g(y)}$ but the principle is exactly the same.

Showing that $f(x,y)=\dfrac{x}{y}$ is continuous when $x>0$ and $y>0$.

So the possible discontinuity points (if any) are these which annulate the denominator.

In your case $d(x,y)=x^2+y^2-1=0$, represents a circle $\mathcal C$ of radius $1$.

But it is only a necessary condition, it is possible that not all the points of $\mathcal C$ (or even none of them) are discontinuity points.

Note that for $(x_0,y_0)\in\mathcal C$ then if $n(x_0,y_0)\neq 0$ then $f(x_0,y_0)$ is an infinite quantity, we say that $(x_0,y_0)$ is a pole of $f$.

But what happens if $n(x_0,y_0)=0$ too ? In this case, we have an undetermined form $\dfrac 00$ and have to study deeper if the ratio has a limit.

In the case where the limit exists: $\lim\limits_{(x,y)\to(x_0,y_0)}f(x,y)=\dfrac{n(x,y)}{d(x,y)}=C\neq\infty$ we can extend $f$ by continuity in this point $(x_0,y_0)$.

If the limit is infinite or does not exist (i.e $f$ has multiple limits depending on the path chosen) then $f$ is not continuous at $(x_0,y_0)$.

Eventually, the discontinuity points are only a subset of $\mathcal C$.

So let's get back to study first if there are points of $\mathcal C$ for which $n(x,y)=0$.

A point on $\mathcal C$ can be represented by $(\cos(t),\sin(t))$ and $n(x,y)=0\iff 2x=5y\iff 2\cos(t)=5\sin(t)\iff \tan(t)=\dfrac 25$

So if we call $\alpha=\tan^{-1}(\frac 25)$ then there are two points on the circle where $f$ might not be discontinuous $(\pm\cos(\alpha),\pm\sin(\alpha))$.

Yet let's set $\begin{cases} (x,y)=(r\cos(t),r\sin(t))\to(x_0,y_0)\\r=\pm(1+u)\quad u\to 0\\t=\alpha+v\quad v\to 0\end{cases}$

When substituting in $f$ and develop we arrive to $f(x,y)\sim k\ \dfrac{\sin(v)}u$ where $k$ is a constant depending of $\alpha$.

This has no limit, in general, when $(u,v)\to(0,0)$ thus $f$ is also not extendable by continuity at the two candidate points.

Finally, $f$ is not defined and cannot be extended by continuity at any point of $\mathcal C$, and since it is defined and continuous elsewhere, we say that $f$ is discontinuous on whole $\mathcal C$.

zwim
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Yes, it does.

The general definition of continuity is this:

Let $f:X\to Y$ be a function defined on a topological space $X$, with values in a topological space $Y$. $f$ is said to be continuous at the point $x\in X$ if for every open set $V\subset Y$ containing $f(x)$, the set $f^{-1}(V)$ is open in $X$.

Since the definition uses the value $f(x)$, it is tacitly assumed that $f$ is defined at $x$. Therefore, in first-year calculus courses, the definition of continuity at a point actually emphasizes the fact that $f$ is defined at the point of continuity.

uniquesolution
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