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The equation is :

$$f(x,y) = \frac{x^2 + y^2 } {x^2 + y^2} $$

I understand that a function is continuous if limit as $(x,y)$ approaches $(a,b)$ of $f(x,y) = f(a,b)$ but I am still a bit confused as to the answer. I know when plugging in $(0,0)$ for x,y the answer is $\frac{0}{ 0}$ but does this mean that the answer is 0- so yes it is continuous, or undefined (since its' dividing by zero), so it is not continuous?

This is for my Vector Calculus/ Multi-variable calculus course.

I'd appreciate any input, thank you!

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3  
$f(x,\ y)$ is undefined at $(0,\ 0).$ –  M. Strochyk Jun 6 '13 at 16:07
3  
As @M.Strochyk says, it is undefined. This is a case where we can define $f$ at $(0,0)$ so that $f$ is continuous there, but as it stands, it is not defined. –  Thomas Andrews Jun 6 '13 at 16:10
    
@ThomasAndrews This function is not continuous at $(0,0)$, to be continuos, $f(0,0)$ should be defined. Although, you can always do a prolongation by continuity to let $f$ be defined at that point and don't change the other values. If you want to do so, we $f(x,y)=1$. It only then that the function becomes continuous. –  moray95 Jun 6 '13 at 16:17
1  
@moray95 I'm sure you just enlightened him, judging by his comment, it doesn't look like he knew that at all. –  Git Gud Jun 6 '13 at 16:21
    
@moray95 I didn't say it was continuous. Read what I read. It can be extended to be continuous, but it is not continuous. –  Thomas Andrews Jun 6 '13 at 16:36

2 Answers 2

up vote 2 down vote accepted

This function cannot be continuous at $(0,0)$ because it's undefined there. Recall that a function $f : A\subset \Bbb R^2 \to \Bbb R$ is continuous at $(0,0) \in \Bbb R^2$ if given $\varepsilon > 0$ there is $\delta > 0$ such that:

$$\sqrt{x^2+y^2} < \delta \Longrightarrow |f(x,y) - f(0,0)|<\varepsilon$$

Now, notice that in your case $f(0,0)$ is undefined, so that this statement doesn't make sense. Every time a function is undefined in some point it cannot be continuous there because this kind of statement won't make sense. In other words, $(0,0) \notin A$ and it's clear from the definition that a function can only be continuous at a point of it's domain.

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And just for fun: it isn't true that it is discontinuous at $(0,0)$. –  Git Gud Jun 6 '13 at 16:42

As noted in the comments this function is not defined at $(0,0)$ because we would be dividing by zero. You can however ask whether $\lim_{(x,y)\rightarrow(0,0)}(f(x,y))$ exists. To show that the limit does exist observe that for any $(x,y)\neq (0,0)$ that:

$f(x,y)=\frac{x^2+y^2}{x^2+y^2}=1$ (notice I did not divide by $0$)

Hence the limit as $(x,y)\rightarrow(0,0)$ does exist and it equals 1. Therefore you can extend this function to a continuous function $\tilde{f}(x,y)=\begin{cases} f(x,y) &\text{if }(x,y)\neq(0,0)\\ 1&\text{if } (x,y)=(0,0) \end{cases}$

But as your question stands, the answer is no.

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And so we have "removed" the discontinuity with...a piecewise function. THANK YOU!!!!!!!! –  imranfat Jun 7 '13 at 16:51

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