Preface: I took calculus back during my undergrad years but never took real analysis (the plight of taking an engineering major). I am currently reading up on analysis concepts in order to shore up several foundational gaps that have tormented me. As of the time I am posting this, I think I have a good grip on epsilon-delta limits of single-variable functions/sequences.
One part of the Wikipedia article (https://en.wikipedia.org/wiki/Limit_of_a_function#Pointwise_limits_and_uniform_limits) for limits of a multivariable function confuses me: the article claims that $f(x,y) = \frac{x}{\cos(y)}$ has a pointwise limit of $0$ for every value of $y$ [emphasis mine own] as $x$ tends to $0$. To be more specific about the notation:
$\lim_{x\to0} \frac{x}{\cos(y)}=0 \qquad \forall y\in \mathbb{R}$
I find the claim that the limit exists at every value of $y$ to be suspect, because if I try to deconstruct this claim with epsilon-delta, that would mean that for any given value of $y$, I need to find a neighborhood around $x=0$ such that the following inequality holds:
$|\frac{x}{\cos(y)}|< \epsilon \Leftrightarrow -\epsilon < \frac{x}{\cos(y)}<\epsilon$
This is where problems arise: the range of $\cos(y)$ is $[-1,1]$, which means that for any $y$ that is equal to $\pm \frac{\pi}{2}\pm2\pi N$, $\cos(y)$ will equal zero, which would cause $\frac{x}{\cos(y)}$ to end up being division by zero. For all other values of $y$, $\cos(y)$ will obviously not be zero, and I can easily understand that the limit exists and would equal zero, but I don't see why the limit would exist at values of $y$ where $y = \pm \frac{\pi}{2}\pm2\pi N$, and the Wikipedia article does not explain why.
If this Wikipedia article's claim is correct, what am I missing?