Multivariable calculus - Limits Show that the function 
$f(x,y)=\frac{y}{x-y} $ for $ x\to 0 $, $ y\to 0 $ can take any limit.
Construct the sequences $ \{f(x_n, y_n\}$ with $ f(x_n, y_n)\to(0,0)$ in such way that the $ \lim_{n\to ∞} f(x_n, y_n)$ is $ 3, 2, 1, 0, -2 $.
Hint: $ y_n=kx_n $.
I am not sure if I am right, but I did the following:
$ f(x,y)=\frac{kx_n}{(x_n-kx_n)} = \frac{kx_n}{x_n(1-k)} = \frac{k}{1-k}$
$ \frac{k}{1-k}=A$
After a few steps I have obtained:
$ k=\frac{A}{(1+A)}, A≠-1$
So, the function can take any limit except -1.
Now we have:  $ A \{3, 2, 1, 0, -2\}$
If $ A=3 $:
$k=\frac{3}{4}, y_n=\frac{3}{4}x_n $
So $ \lim_{n\to ∞} \frac{\frac{3}{4}x_n}{\frac{1}{4}x_n}=3$.
For $A=2, 1, 0$ and $-2$ I did the same.
I am not sure if this is the end of the exercise. I don't understand what exactly in this context is meant by the sequence.
I also don't understand why $x_n= \frac{1}{n}\to0$ and  $y_n= \frac{k}{n}\to0$ is. How can I obtain $ x_n=\frac{1}{n}\ $ and $ y_n=\frac{k}{n}$ ?
Any help is appreciated.
Thanks in advance.
 A: If $(x_n)$ is a sequence converging to $0$ and $y_n=kx_n$ then you have
$$f(x_n,ky_n)=\dfrac{k}{1-k}.$$ (Obviously we need $x_n\ne 0$ and $k\ne 1.$ In other case, $f(x_n,ky_n)$ is not defined.) This shows that $\lim_{(x,y)\to (0,0)} f(x,y)$ doesn't exist because for different $k$'s you get different values. 
Note that $$\lim_{(x,y)\to (0,0)} f(x,y)=\lim_{(x_n,y_n)\to (0,0)} f(x_n,y_n)$$ for all sequences $(x_n),(y_n)$ converging to $0$ in case the limit exists. Since the RHS has different values for different sequences you can conclude that $\lim_{(x,y)\to (0,0)} f(x,y)$ doesn't exist.
It is not necessary to get the possible values of $\lim_{(x_n,y_n)\to (0,0)} f(x_n,y_n).$ If you have two pairs of sequences and the limit takes two different values you can conclude that $\lim_{(x,y)\to (0,0)} f(x,y)$ doesn't exist. But you are asked for it. Indeed the question says:

Construct the sequences $ \{f(x_n, y_n\}$ with $ f(x_n, y_n)\to(0,0)$ in such way that the $ \lim_{n\to ∞} f(x_n, y_n)$ is $ 3, 2, 1, 0, -2 $.
  Hint: $ y_n=kx_n $.

Your construction of the sequences is correct.
