Evaluate $\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $ This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.
$$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$
My attempt: 
Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):
$$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$ 
If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?

We want to show that $\forall \epsilon \gt 0 : \exists \delta \gt 0 : ||(x,y)||< \delta \implies |\frac{\ln(1-x-y)}{x+y}+1| \lt \epsilon$.
Fix $\epsilon \gt 0$. We know that $\lim_{\phi \to 0} \frac{ln(1+\phi)}{-\phi} = -1$. So there's $\delta_1 \gt 0$ such that $|\phi| \lt\delta_1 \implies |\frac{ln(1+\phi)}{-\phi}+1| \lt \epsilon$. Let $\delta = \delta_1$ and $\xi = -x-y$. Suppose $||(x,y)|| \lt \delta $. Since all norms in $R^n$ are equivalent we can use the norm of the sum. Then:
$||(x,y)|| = |x|+|y| \geq |x+y| = |\xi|$. Since $|\xi| < \delta_1 $, then $|\frac{ln(1+\xi)}{-\xi}+1| < \epsilon$ therefore $|\frac{ln(1-x-y)}{x+y}+1| < \epsilon  $. $\square$
