Limit of a multivariable piecewise function 
Let $f:\mathbb{R^2}\to\mathbb{R}$ given by:
$f(x,y) =
  \begin{cases} 
      x-y+1     & \text{if }xy\geq 0 \\
      y-x-1  & \text{if } xy<0 \\
  \end{cases}$

I need to compute $\lim_{(x,y)\to(0,1)}f(x,y)$, but I'm not sure about the following reasoning:
The problem here is that $f$, close to $(0,1)$, corresponds to different formulas.
If $D_1=\{(x,y)\in\mathbb{R^2}: xy\geq0\}$ and $D_2=(D_1)^C$, we have
$\lim_{(x,y)\to(0,1),(x,y)\in D_1}f(x,y)=0$ and the same for $(x,y)\in D_2$, because $f$ is continuous on $D_1$ and $D_2$. So, can I conclude that the limit is $0$?
My english isn't very good; if you find any errors, please correct them.

What you've done seems generally okay, yes, except for one minor point: the fact that $f$ is continuous on $D_{2}$ is irrelevant since $(0,1)\not\in D_{2}$. What you mean to say is that you've simply evaluated the limit using the expression that gives the values that $f$ takes on $D_{2}$, which should be perfectly acceptable here.
The following hint is intended to provide a slicker method.
Hint: notice that $x-y+1=-(y-x-1)$. Now, evaluate the limit $\lim_{(x,y)\to(0,1)}\lvert f(x,y) \rvert$. What can you conclude?
