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.

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,
$$\begin{align}\lim_{x\to 0^+} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=+\infty\\ \lim_{x\to 0^-} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=-\infty\end{align}$$
