How to prove that $x/y$ is continuous in R $f:R^2$ \{y=0} $\Rightarrow R$ , $f:(x,y)\Rightarrow x/y$.
Prove (formally) that $f$ is continuous. 
I think what I should show is that any point that belongs to an open ball of radius $\epsilon$ of image, has a pre-image that belongs to an open ball around (x,y), and since image and pre-image are both open, then $f$ is continuous. But this doesn't seem correct to me.
Any help is appreciated.  
 A: Yes, to show that such a function $f$ is continuous you want to show that the preimage of an open ball of radius $\epsilon$ contains an open ball around $(x,y)$. Here's why you should believe this is true. If $(x,y)$ is close to $(x',y')$, say 
$$\sqrt{(x' - x)^2 + (y'-y)^2} <\delta$$
then we know that $|x-x'|<\delta$ and $|y-y'| < \delta$ for some small $\delta>0$. Try and compute
$$\left| \frac{x}{y} - \frac{x + \eta}{y +\rho} \right|$$
For $|\eta|,|\rho| < |\delta|$. Can you show that this can be made less than $\epsilon$ by making $\delta$ small enough?
A: You can use limits to show it if you want or open balls. Using limits, the following facts is easy: The quotient of two continuous functions that go to $\mathbb{R}$ is continuous, wherever the functions are defined and the denominator is not 0.
After proving that, it is easy to see that the projections $\pi_1:(x,y)\mapsto x$ and $\pi_2:(x,y)\mapsto y$ are continuous, since if $B_r(x)$ is an open ball of radius $r$ around $x$, then $\pi_1^{-1}(B_r(x))=B_r(x)\times\mathbb{R}$, which is an open set in $\mathbb{R}^2$. We then have that $x/y$ is the quotient of two continous functions: if $\phi(x,y)=x/y$, then $\phi(x,y)=\pi_1(x,y)/\pi_2(x,y)$, and so is continuous.
