What is $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y}$? I want to understand this multivariate limit.
WolframAlpha says $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} = 0$
But what if I take a curve $y = -x$, then the limit doesn't exist, right? Wouldn't this make the limit inexistent?
I tried to prove using sandwich theorem and definition, but didn't get anything good.
Any help would be appreciated.
 A: The function $f(x, y) = \frac{1-\cos(x)}{x+y}$ is not defined if $x+y = 0$. But even if restricted to the domain
$$
D = \{ (x, y) \in \Bbb R^2 \mid x+y \ne 0 \} \,
$$
the limit does not exist: For $a \ne 0$ we have $(x, ax^2-x) \to (0, 0)$ for $x \to 0$, but the limit
$$
 \lim_{x \to 0} f(x, ax^2-x) =  \lim_{x \to 0}\frac{1-\cos(x)}{ax^2} = \frac{1}{2a}
$$
depends on $a$.
A: Method $-1.$
$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=2\lim_{(x,y) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}\end{align}$$
Let  $y(x):=r\sin^2\left(\frac x2\right)-x$, as $x\to 0$ and $r\in\mathbb R\setminus \left\{0\right\}$ then we get

$$\begin{align}2\lim_{(x,y(x)) \to (0,0)} \dfrac{\sin^2(\frac x2)}{x+y}=2\lim_{x\to 0}\frac {\sin^2\left(\frac x2\right)}{r\sin^2\left(\frac x2\right)}=\frac 2r\end{align}$$

This means, the limit doesn't exist. Because, the evaluation of the original limit is dependent on $r.$

Method $-2.$
$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=2\lim_{(x,y) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}\end{align}$$
Let  $y(x):=\left(\frac x2\right)^3-x$ as $x\to 0$, then we have

$$\begin{align}2\lim_{(x,y(x)) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}=2\lim_{x\to 0} \dfrac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}\longrightarrow \text{limit does not exist.}\end{align}$$

Because,
$$\begin{align}\lim_{x\to 0^+} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=+\infty\\ \lim_{x\to 0^-} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=-\infty\end{align}$$
A: $$
\begin{aligned}
\lim\limits_{(x, y) \rightarrow (0, 0)}\frac{1-\cos(x)}{x + y} &= 
\left|
\begin{array}{c}
\text{changing the variables:} \\
x = r\cos(\theta) \\
y = r\sin(\theta) \\
(r \in (0, +\infty), \text{ }\theta \in [0, 2\pi]) \\
\Downarrow \\
(x, y) \rightarrow (0, 0) \Leftrightarrow r \rightarrow 0
\end{array}
\right| = 
\lim\limits_{r \rightarrow 0}\frac{1-\cos\left(r\cos(\theta)\right)}{r(\cos(\theta) + \sin(\theta))} = \\
=\lim\limits_{r \rightarrow 0}\frac{2\sin^2\left(\frac{r}{2}\cos(\theta)\right)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} &= 
\left|
\begin{array}{c}
\sin^2\left(\frac{r}{2}\cos(\theta)\right) \underset{r \rightarrow 0}{\sim} \frac{r^2}{4}\cos^2(\theta)
\end{array}
\right| = 
\lim\limits_{r \rightarrow 0}\frac{\frac{r^2}{2}\cos^2(\theta)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} = \\
\frac{\cos^2(\theta)}{2\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)}\lim\limits_{r \rightarrow 0}r = 0, \forall \theta\text{ }\left(\theta \ne \frac{3\pi}{4}\right).
\end{aligned}
$$
So the limit equals to zero for all $\theta$ except $\theta = \frac{3\pi}{4}$.
Now, let us check the case when $\theta \rightarrow \pm\frac{3\pi}{4} \Leftrightarrow y \rightarrow -x$.
In this case,
$$
\lim\limits_{\theta \rightarrow \frac{3\pi}{4}}\lim\limits_{r \rightarrow 0}\left(\cdot \right) = 0 \ne \pm\infty = \lim\limits_{r \rightarrow 0}\lim\limits_{\theta \rightarrow \pm\frac{3\pi}{4}}\left(\cdot \right)
$$
This means that the limit doesn't exist.
