TITLE: Determine whether $\lim_{(x,y)\to (2,-2)} \frac{\sin(x+y)}{x+y}$ exists.
QUESTION [11 upvotes]: I am trying to determine whether $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ exists. I should be able to use the following definition for a limit of a function of two variables:

Let $f$ be a function of two variables that is defined on some open disk $B((x_0,y_0),r)$, except possibly at the point $(x_0,y_0)$ itself. Then
$$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L$$
if for any $\varepsilon>0$, however small, there exists a $\delta>0$ such that
$$\text{if } 0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta \text{ then } |f(x,y)-L|<\varepsilon$$

The limit definition above applied to the limit $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ requires that the function $\dfrac{\sin(x+y)}{x+y}$ be defined on every point of some open disk $B((2,-2),r)$ (with the possible exception of the point $(2,-2)$).
My problem here is that the function $\dfrac{\sin(x+y)}{x+y}$ is undefined for $x+y=0$, that is, it is undefined for all points in the line $y=-x$. If I understood it correctly, it seems to mean that I can't find any open disk $B((2,-2),r)$ on which the given function is defined. Therefore, it seems that I can't apply the limit definition.
What am I misunderstanding here?
The argument above makes me believe that this limit does not exist (because I can't apply the definition), but Wolfram|Alpha says that the value of this limit is $1$. Why this apparent contradiction?
Edit: I should point out that this problem and the limit definition that I'm using are both taken from the same book. So, the book assumes that I will use the given limit definition.
Update
I have one more doubt concerning this.
The same book that I'm using says that the function below is continuous for all points $(x,y)$ in $R_2$.
$$f(x,y)=\begin{cases} \dfrac{\sin(x+y)}{x+y} & \text{ if $x + y \neq 0$} \\ 1 & \text{ if $x + y = 0$}\end{cases}$$
The proof is as follows:

Let $h(t)=\begin{cases} \dfrac{\sin(t)}{t} & \text{ if $t \neq 0$} \\ 1 & \text{ if $t = 0$}\end{cases}$ and $g(x,y) = x+y$; so, $f(x,y) = h(g(x,y))$. The domain of $g$ is $R_2$ and it is continuous everywhere in its domain. Since $g(x,y)\to 0$ as $x+y\to 0$ and $h$ is continuous at $0$, $h(g(x,y))$ is continuous everywhere.

The definition of continuity in this book states that, for a two-variable function $g$ to be continuous at $(x_0,y_0)$, $g(x_0,y_0)$ has to exist and $\lim_{(x,y)\to (x_0,y_0)} g(x,y)$ has to be equal to $g(x_0,y_0)$. So, by the given definition of continuity, $f(x,y)$ being continuous everywhere means that $\lim_{(x,y)\to (x_0,y_0)} f(x,y) = f(x_0,y_0)$ for all $(x_0,y_0)$.
The proof above seems to imply that $\lim_{(x,y)\to (2,-2)} \dfrac{\sin(x+y)}{x+y}$ should exist. But the problem is that this limit should not exist, considering the definition of limit that I am using in this question. Is there a contradiction here or am I missing something?

REPLY [2 votes]: Using the bona fide interpretation of the expression
$$f(x,y):={\sin(x+y)\over x+y},$$
it is defined on the set $D:={\mathbb R}^2\setminus\{(x,-x)\ |\ x\in{\mathbb R}\}$. The point $(2,-2)$ is an accumulation point of $D$; therefore it is allowed to consider the $\lim_{(x,y)\to(2,-2)} f(x,y)$. Here it is tacitly understood that $f$ is only evaluated in points of $D$. 
The function
$${\rm sinc}:\quad {\mathbb R}\to{\mathbb R},\qquad u\mapsto\cases{{\sin u\over u}\quad &$(u\ne0)$ \cr \ 1&$(u=0)$\cr}$$
is well known to be continuous on all of ${\mathbb R}$; therefore the function
$$\tilde f(x,y):={\rm sinc}(x+y)$$
is continuous on all of ${\mathbb R}^2$. Since for $(x,y)\in D$ we have $f(x,y)=\tilde f(x,y)$ it follows that
$$\lim_{(x,y)\to(2,-2)} f(x,y)=\lim_{(x,y)\to(2,-2)}\tilde f(x,y)=\tilde f(2,-2)=1\ .$$