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

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

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?

• Did you try letting $(x+y) = u$ and taking $\lim_{u\to0} f(u)$? And this limit is 1 by the way. May 10, 2015 at 5:00
• Why is it reasonable to make that substitution? May 10, 2015 at 5:04
• @getaflix see my answer. May 10, 2015 at 5:05
• The substitution is reasonable because no matter how $x+y$ changes, it changes in exactly the same way everywhere. May 10, 2015 at 5:07

Let $x+y=t$ so that $t\to 0$ as $(x, y) \to (0,0)$ therefore we have that $$\lim_{(x, y) \to (0,0)}\frac{\sin (x+y)}{x+y}=\lim_{t \to 0}\frac{\sin t}{t}=1$$
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 $\lim_{u\to0} \frac{\sin u}{u} = 1$.
• @getafix substitution is a fancy way of making it simple. You can also perform the same trick without substitution, just use $x+y$ instead. May 10, 2015 at 5:07
• The substitution technique is not a valid proof. It prove the limit exists as $x+y$ approaches 0, not $(x,y) \rightarrow (0,0).$ See an earlier post for a better proof. \math.stackexchange.com/questions/291538/… May 10, 2015 at 5:13
$$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ 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 $$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\frac{(x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\left[1-\frac16 (x+y)^2+O\left((x+y)^4\right)\right]=1$$