Study continuity of this function Hello im studying calculus at the university and I dont know how to solve the following exercise:
Study the continuity of the next function:
$$f(x,y) = \begin{cases} \frac{x^2-xy}{x+y}&\text{for } x+y\ne0\\ 0 &\text{for }(x,y) =(0,0). \end{cases}$$
I've tried to resolve it with iterated limits and directional limits, but im sure if its correct.
 A: For the function to be continuous at a point $(x_0,y_0)$, you need to prove that
$$ \lim f(x,y) = f(x_0,y_0)\quad \mathrm{as}\quad (x,y)\to (x_0,y_0). $$
To find the limit of the function at the point $(0,0)$, use the polar coordinates $x=r\cos(\theta), y= r\sin(\theta)$ and consider taking the limit as $r\to 0.$
A: mrf's solution is simple and correct. Here I want to show an alternative one.
The only question is $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ or not.
$$
f(x,y)=\frac{x^2-xy}{x+y}=x\frac{x+y-2y}{x+y}=x-2\frac{1}{\frac{1}{x}+\frac{1}{y}}.
$$
Obviously $\lim_{(x,y)\to(0,0)}x=0$. If we can choose $y:=y(x)$, ($y(x)\neq -x$), such that $\lim_{x\to 0}\left(\frac{1}{x}+\frac{1}{y(x)} \right)\neq\pm\infty$, then $f$ is not continuous in $(0,0)$. Choose, for example, $\frac{1}{x}+\frac{1}{y(x)}=1$, that is, $y(x):=\frac{x}{x-1}$, where we may assume that $x\neq 1$ because $x\to 0$. ($y(x)\neq -x$ for any $x\neq 0$.)
A: Note that $f(0,t) = 0$ (for $t \neq 0$), but $$f(t,-t+t^2) = \frac{t^2+t^2-t^3}{t^2} = 2-t,$$ again for $t \neq 0$. What happens when $t \to 0$? This shows that $f$ is not continuous at $(0,0)$. 
On the other hand, $f$ is continuous everywhere else where $f$ is defined, since the numerator and denominator clearly are continuous. 
A: $$ \text{We choose a path like} \hspace{20mm} y=mx$$
$$ f=\frac{x^2-xy}{x+y}=\frac{x^2-x^2m}{x+my}=\frac{x^2(1-m)}{x(1+m)}$$ 
$$ \lim_{x \rightarrow \alpha} \frac{x(1-m)}{(1+m)}=\frac{\alpha (1-m)}{(1+m)}$$
$$\text{answer of the limit is depend on path (m) , so } \mathcal{f} \text{ is not continuous} $$
A: Yet another way to show non-continuity in $(0,0)$ is to show that $f$ is unbounded in every neighborhood of $(0,0)$: 
Let $\epsilon > 0 $ and $x\in\mathbb{R}$ such that $0<x<\frac{\epsilon}{2}$. Let $\delta\in\mathbb{R}_+$. We have
$$\|(x, -x + \delta)\|_2 = \sqrt{x^2 + (-x+\delta)^2} = \sqrt{2x^2 - 2x\delta + \delta^2}\\
\leq \sqrt{2x^2 + \delta^2} \leq \sqrt{\frac{\epsilon^2}{2} + \delta^2} < \epsilon
$$
for sufficiently small $\delta$, so $(x, -x + \delta)\in B_{\epsilon}(0,0)$. However,
$$ f(x, -x + \delta) = \frac{x^2- x(-x+\delta)}{\delta} = \frac{2x^2}{\delta} - x$$
and we see that
$$ \lim_{\delta\rightarrow 0}\; \left| f(x,-x+\delta) \right| = \infty, $$
i. e. $f$ is unbounded on $B_{\epsilon}(0,0)$. 
