is f(x,y) continuous at (0,0)? 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!
 A: 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.
A: 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.
