<s>[INST] 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.
 [/INST] $$
\begin{aligned}
\lim\limits_{(x, y) \rightarrow (0, 0)}\frac{1-\cos(x)}{x + y} &= 
\left|
\begin{array}{c}
\text{changing the variables:} \\
x = r\cos(\theta) \\
y = r\sin(\theta) \\
(r \in (0, +\infty), \text{ }\theta \in [0, 2\pi]) \\
\Downarrow \\
(x, y) \rightarrow (0, 0) \Leftrightarrow r \rightarrow 0
\end{array}
\right| = 
\lim\limits_{r \rightarrow 0}\frac{1-\cos\left(r\cos(\theta)\right)}{r(\cos(\theta) + \sin(\theta))} = \\
=\lim\limits_{r \rightarrow 0}\frac{2\sin^2\left(\frac{r}{2}\cos(\theta)\right)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} &= 
\left|
\begin{array}{c}
\sin^2\left(\frac{r}{2}\cos(\theta)\right) \underset{r \rightarrow 0}{\sim} \frac{r^2}{4}\cos^2(\theta)
\end{array}
\right| = 
\lim\limits_{r \rightarrow 0}\frac{\frac{r^2}{2}\cos^2(\theta)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} = \\
\frac{\cos^2(\theta)}{2\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)}\lim\limits_{r \rightarrow 0}r = 0, \forall \theta\text{ }\left(\theta \ne \frac{3\pi}{4}\right).
\end{aligned}
$$
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.
</s>