Question about continuity of function with two variable Given $f = xy^{1/2}$ I am supposed to find 
if $f_y$ is continuous at $(0,0)$ or not.
I know that $f_y = \frac12 xy^{-1/2}$
is not continuous at $(0,0)$,
because if I take two paths $x=0$ and $y=x^2$
to $(0,0)$, then I get two different values.
I was just wondering is it possible to conclude
that it's not continuous just because
$f_y = \frac12 xy^{-1/2}$
is not well defined 
(i.e. the 0/0 form) at $(0,0)$?
 A: Yes. A function is continuous at a point $\mathbf{x}_0$ iff $\lim_{\mathbf{x} \rightarrow \mathbf{x}_0} f(\mathbf{x}) = f(\mathbf{x}_0)$.
A: Indeed, there are two ways to prove the lack of continuity:


*

*show that $f_y(x,y)=\frac12 xy^{-1/2}$ when $y>0$, and argue that $\lim_{(x,y)\to (0,0)}xy^{-1/2}$ does not exist. 

*show that the derivative $f_y$ does not exist at $(0,0)$. If a function is not defined at a point, it cannot be  continuous at that point.


You implemented approach 1 correctly. But I do not like your treatment of approach 2. The formula $f_y(x,y)=\frac12 xy^{-1/2}$ is valid only under the assumption $y> 0$, because it was obtained (from differentiation rules) using that assumption. When $y=0$, you don't have that formula. So the fact that $ xy^{-1/2}$ is undefined at $(0,0)$ does not tell you that $f_y$ is undefined at $(0,0)$. 
The correct way to implement approach 2 is to use the definition of $f_y(0,0)$ as the limit
$$\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}$$ and argue that this limit does not exist.
