Basic Multivariable Limit The other day I was asked to solve the following multivariable limit:
$$\lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}$$
As simple as it was I was stumped. I tried some algebraic manipulation using conjugates, which got me nowhere. Then I tried defining $x$ and $y$ by their polar definitions and got stuck here:
$$\lim_{r\to0}\frac{r\cos\theta-r\sin\theta}{r\cos\theta+r\sin\theta}$$
$$=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}$$
But I didn't know what to do from here.
The only other approach that struck me as a possibility was partial fractions.
What's the best way to tackle this limits problem and ones like it?
 A: This limit does not exist. If you approach $(0,0)$ through the $x$-axis, you get 
$$\lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}=\lim_{x\to0}\frac{x}{x}=1,$$
But if you approach it by the line $y=x$, you get
$$\lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}=\lim_{x\to0}\frac{x-x}{x+x}=0,$$
So the limit can't exist.
A: Your “polar coordinates” method can work:
$$\lim_{r\to0}\frac{r\cos\theta-r\sin\theta}{r\cos\theta+r\sin\theta}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}$$
If the two-variable limit existed, the expression on the right would not depend on $\theta$. But clearly it does depend on $\theta$.  If $\theta = 0$, the limit is $1$.  But if $\theta = \frac{\pi}{2}$, the limit is $-1$.  
A: Take the next sequences: 
$x_k=\left\{\displaystyle\frac{1}{k},\displaystyle\frac{1}{k} \right\}$. Clearly, $\lim\limits_{k\to\infty}x_k=(0,0)$. And, $\lim\limits_{k\to\infty}f(x_k)=0$
$y_k=\left\{ \displaystyle\frac{1}{k},0\right\}$. Clearly, $\lim\limits_{k\to\infty}y_k=(0,0)$ and $\lim\limits_{k\to\infty}f(y_k)=1$
We can conclude that the limit doesn't exist. Why? We take two sequences such that converges to $(0,0)$ but the sequence of images converges to two different values.
