Limit of $\frac{\sin(x+y)}{x+y}$ as (x,y)→(0,0)

Question

$$ \lim\limits_{(x, y)\to (0, 0)}\frac{\sin(x+y)}{x+y} $$

I did the following

$a)$ along $x$ axis, the limit is one

$b)$ along $y$ axis the limit is one

$c)$ along $y=x$ the limit is one

Since there exists more ways to approach the origin, I know I cannot conclude from the steps given above.

$d)$ along $y= -x$

$\frac{\sin(x-x)}{x-x}$ is not defined. Isn't it still possible for the function to have a limit, even though it is not defined at that point?

How do I conclude whether a limit exists or doesn't in such a case?

EDIT

So for, $$ \lim\limits_{(x, y)\to (0, 0)}\frac{\sin(xy)}{xy} $$ Can I proceed in a similar manner and perform a substitution and state the limit is 1?

Answer 1

So $\lim_{(x,y) \to (0,0)} \frac{\sin(x+y)}{x+y} = 1$, since this is a simple trig limit and the form is $\frac{\sin(0)}{0}$. You might substitute $u = x+y$, and $\lim_{(x,y)\to(0,0)} x+y$ is obviously 0. Doing the substitution, we get $\lim_{u\to0} \frac{\sin u}{u} = 1$.

Answer 2

Let $x+y=t$ so that $t\to 0 $ as $(x, y) \to (0,0)$ therefore we have that $$\lim_{(x, y) \to (0,0)}\frac{\sin (x+y)}{x+y}=\lim_{t \to 0}\frac{\sin t}{t}=1$$

Answer 3

$$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ The Taylor series of $\sin(x+y)$ at around $(0, 0)$ is $$ (x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right) $$ Therefore $$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\frac{(x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\left[1-\frac16 (x+y)^2+O\left((x+y)^4\right)\right]=1$$