Multivariable continuity $f(x,y) = \frac{\sin x \ +\ \sin y}{x\ +\ y}$

Question

How do you show that the following multivariable function is continuous?

$f(x,y) = \frac{\sin x + \sin y}{x +y}$

I think I want to show that for every point $(x_0,y_0)$:

$\forall \varepsilon >0, \ \exists\delta >0$ such that $||(x,y)-(x_0,y_0)|| < \delta \implies |f(x,y) - f(x_0,y_0)| < \varepsilon $

I know that the function is not continous at $x = - y$ as the denominator is undefined then.

$\sin x$ and $\sin y$ are continuous on $\mathbb{R}$ as is $\frac{1}{x+y}$ (except for x = -y)

Answer 1

Your function is continuous on $\mathbb{R}-D$, where $D$ is the diagonal $ \left\lbrace (x,-x), x \in \mathbb R\right\rbrace$, since the numerator and the denominator are continuous, and the denominator is not zero.

Let $a=(x,-x)$ and $a_n$ a sequence of $\mathbb{R}^2$ going to $a$. Let $a_n=(x_n,y_n)$.

$$ \sin(x_n)+\sin(y_n)=2\sin((x_n+y_n)/2)\cos((x_n-y_n)/2) \sim 2 \dfrac{x_n+y_n}{2}\cos((x_n-y_n)/2) $$

since $x_n+y_n \to 0$

Consequently $f(x_n,y_n) \sim\cos((x_n-y_n)/2) \to_{n\to\infty} \cos(x)$.

If you define $f(x,-x)=\cos(x)$ then your function is continuous at $a$.

And finally $f$ is continuous on $\mathbb{R}^2$.

Answer 2

Change the coordinates to $u=(x+y)/2$ and $v=(x-y)/2$, so that $x=u+v$ and $y=u-v$. Then we have

$$f(x,y)={\sin(u+v)+\sin(u-v)\over2u}={(\sin u\cos v+\cos u\sin v)+(\sin u\cos v-\cos u\sin v)\over2u}={\sin u\over u}\cos v\to1\cdot1=1$$

as $(u,v)\to(0,0)$ (along any path with $u\not=0$).