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Given a multivariable function $f(x,y) = \frac{x^2-y^2}{x^2+y^2}$, we know at the point $f(0,0)$, the function is undefined. So if I were to test for continuity, couldn't I say that this function is non-continuous at the point $(0,0)$ because there is a hole?

I ask this because the correct way to show that the function is discontinuous is by showing that along the axis $y=0$, the limit of the function approaches 1. Along the axis $x=0$, the limit approaches -1, so the function must not be continuous.

Why is it/what is the point of showing the limit doesn't exist along different paths, if we already know that the point is undefined?

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  • $\begingroup$ If the limit were the same along all paths, then we could define the value of the function at that point to be said value. $\endgroup$ – Math1000 Oct 5 '19 at 20:33
  • $\begingroup$ @Math1000 Even if that said value does not exist because we are dividing by 0? $\endgroup$ – Jay Oct 5 '19 at 20:35
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Unless you assign a value to $f(0,0)$ (“by hand”, not using the formula) it doesn't make sense to ask if the function is continuous or not at that point. (See discussion here, for example.)

What does make sense to ask is whether you can define $f(0,0)$ so that the function becomes continuous. And in this case you can't, since $\lim_{(x,y)\to(0,0)}f(x,y)$ doesn't exist. That is, the function $$ f(x,y) = \begin{cases} \frac{x^2-y^2}{x^2+y^2},& (x,y) \neq (0,0),\\ C,&(x,y)=(0,0) \end{cases} $$ is discontinuous at $(0,0)$ no matter what $C$ is.

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We say that $f$ is continuous at the point $(x_0,y_0)$ if $$\lim_{(x,y) \to (x_0,y_0)}f(x,y)=f(x_0,y_0).$$ This condition fails if either the limit doesn't exist or $f(x_0,y_0)$ isn't defined. In your case, you know $f(0,0)$ isn't defined, hence, you know $f$ isn't continuous at that point.

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It is the same case for functions in one variable, for example $\frac{\sin 2x}{x}$ is undefined at zero, right? But its limit exists at $0$, and it is $2$ so if you took the function $f(x) = \begin{cases} \frac{\sin 2x}{x} & \text{if $x \ne 0$}\\ 2 & \text{$x=0$} \end{cases}$ then this will be a continuous function.

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  • $\begingroup$ What is the usefulness of the information if we are excluding x = 0? $\endgroup$ – Jay Oct 5 '19 at 20:46
  • $\begingroup$ What information exactly you see useless? Getting the limit? We use it to define $f(0)$ $\endgroup$ – Fareed Abi Farraj Oct 5 '19 at 20:48
  • $\begingroup$ I guess my question is this: since f(0,0) is undefined, are we saying the function only applies for non-zero x and y values? And we use the limit to show what the true value of f(0,0) is (if it exists)? $\endgroup$ – Jay Oct 5 '19 at 20:49
  • $\begingroup$ Yes I think what your saying is right, usually it is stated this way, $f$ is defined for all $x \ne 0$ (i.e. it is undefined for $x=0$) but it has an extension by continuity (only if the limit exists) $\endgroup$ – Fareed Abi Farraj Oct 5 '19 at 20:57
  • $\begingroup$ So in reality, with the example you gave, it is undefined at $x=0$. How does knowing the limit at $x=0$ is 2 help us? $\endgroup$ – Jay Oct 5 '19 at 21:00

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