Limit of $\frac{\sin(x+y)}{x+y}$ as $(x,y) \to (0,0)$ I want to prove that $\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} = 1$
Is it sufficient to say that if $u = x+y$ then as $(x,y) \rightarrow (0,0)$ then $u \rightarrow 0$ and then evaluate $\lim_{u\to 0} \frac{\sin u}{u}$ which is 1 by L'Hôpital's rule?
I feel that this is somehow "cheating" and I would really like to prove it in a more rigorous way, i.e., by using an $\epsilon$ - $\delta$ argument.
I have tried for a long time to use the one variable version and prove that if $0 \lt \sqrt{x^2 + y^2} \lt \delta$ then somehow $0 \lt |x+y| \lt \delta$ and it would follow from the one variable case that $| \frac{\sin(x+y)}{x+y} - 1| \lt \epsilon$. However, because $|x+y| = 0$ when $y = -x$ I am not getting anywhere with this.
Any input on this would be helpful :)
 A: As @julien points out, the function $\frac{\sin(x+y)}{x+y}$ is not defined in a neighbourhood of the origin, so the limit does not exist. But there is a natural extension of the function which is defined on a neighbourhood of the origin, and for which the limit exists and equals 1. (There are lots of extensions, but this one seems most natural to me.) Define $f:\mathbb{R}^2\to\mathbb{R}$ by $$f(x,y)=\left\{
\begin{array}{cl} 
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\ 
1 & x+y=0.\end{array}\right.$$
Then $\lim_{(x,y)\to(0,0)}f(x,y)=1$. An $\epsilon-\delta$ proof relies on the one variable result $$\lim_{u\to 0}\frac{\sin u}{u}=1.$$
That is, given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon.$$
It is helpful to notice that this can be rephrased as follows: given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<2\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon. \qquad (*)$$
We want to prove that given $\epsilon>0$, there exists a positive $\delta$ such that $$0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-1|<\epsilon.$$
So let $\epsilon>0$. Choose $\delta$ so that $(*)$ holds true, and let $0<\sqrt{x^2+y^2}<\delta$. Then $$|x+y|\leq|x|+|y|\leq\sqrt{|x|^2+|y|^2}+\sqrt{|x|^2+|y|^2}=2\sqrt{x^2+y^2}<2\delta.$$ Then by (*) (for $x+y\neq0$) and by the definition of $f$ (for $x+y=0$), we have $|f(x,y)-1|<\epsilon$ as required. 
The same argument works if the function is defined as $g:\mathbb{R}^2\setminus\{(x,y):x+y=0\}\to\mathbb{R}$ with $g(x,y)=\frac{\sin(x+y)}{x+y}$. With a slight amendment, the same result also holds for $$f(x,y)=\left\{
\begin{array}{cl} 
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\ 
h(x) & x+y=0,\end{array}\right.$$
where $h$ is any function defined in a neighbourhood of $x=0$ which has $\lim_{x\to0}h(x)=1$.
A: Presumably, when we talk about $\lim_{(x,y)\to(0,0)}\frac{\sin(x+y)}{x+y}$, the domain of $\frac{\sin(x+y)}{x+y}$ is assumed to be $\mathbb{R}^2\setminus\{(x,y):x=y\}$, otherwise, as you and the others have pointed out, the limit does not exist. Now, given that the issue has been settled, it's perfectly OK to evaluate the limit as $\lim_{u\to0}\frac{\sin u}{u}$. Let $u=x+y$ and $v=x-y$. Recall what is meant by $\lim_{(x,y)\to(0,0)}\frac{\sin(x+y)}{x+y}=L$:

(a) For any $\epsilon>0$, there exists $\delta>0$ such that $\|(u,v)\|<\delta \Rightarrow \left|\frac{\sin(u)}{u}-1\right|<\epsilon$.

And what is meant by $\lim_{u\to0}\frac{\sin(u)}{u}=L$ is this:

(b) For any $\epsilon>0$, there exists $\delta>0$ such that $|u|<\delta \Rightarrow \left|\frac{\sin(u)}{u}-1\right|<\epsilon$.

So, suppose you have shown (b). Then, given any $\|(u,v)\|<\delta$, you will have $|u|<\delta$. Hence by (b), you get (a).
A: Use the fact that $|x+y|\leq \sqrt{2}\sqrt{x^2+y^2}$.
If $\epsilon>0$ and $\delta>0$  are such that $|\dfrac{\sin u}{u}-1|<\epsilon$ for $|u|<\delta$ then 
for $(x,y)$ with $\|(x,y)\|=\sqrt{x^2+y^2}<\dfrac{\delta}{\sqrt{2}}\Longrightarrow$ ...
A: $$\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} $$
$$1) let\ y=0 \ Then \lim_{(x,0)\to (0,0)} \frac{\sin(x)}{x} = 1= L_1$$
$$2)let\ y=0 \ Then \lim_{(0,y)\to (0,0)} \frac{\sin(y)}{y} = 1=L_2$$
$$3)\ let\ y=x \ Then \lim_{(x,x)\to (0,0)} \frac{\sin(2x)}{2x} = 1=L_3$$
$$L_1=L_2=L_3=1$$
