$$\lim_{x \to 0} \frac{\cos x^{-2}}{ \cos x^{-2}}$$
I had nothing better to do than come up with a pathological function. The cos(x) function has an infinite number of zeroes which are constantly spaced. cos(x^2) likewise has an infinity of zeroes, but the spacing between them decreases as one goes out to either infinity. cos(x^-2) maps that infinity of zeroes into a finite interval, namely [-1,1]. I understand my quest for a pathological function might end here, but I felt the fact that its limit at zero didn't exist was more intuitive, since it had an infinite number of waves within any interval around x=0.
So anyway, to generate a flat line with a infinite number of gaps on a finite interval around zero, I simply divided cos(x^-2) by itself to get cos(x^-2)/cos(x^-2)
Clearly
$${cos(x^{-2})\over{cos(x^{-2})}}=1;x \neq \sqrt[\scriptstyle-2]{{\pi\over2}+2\pi k} \enspace or \enspace \sqrt[\scriptstyle-2]{{3\pi\over2}+2\pi k}, \enspace k \in \mathbb{N} $$
I asked my classmates about this limit, and they thought little of it, even when I warned them about its odd neighborhood behavior. They were all convinced it was 1, since the function is 1 "almost everywhere" near x=0.
I'll be honest, I'm not even sure if it exists. This pathology prevents one from defining f for an interval of any sort around 0, which I thought might preclude the existence of a limit because you lose some oppurtunities to find a delta within for every epsilon within an interval, since an infinity of epsilons is missing.
I looked at wikipedia's (ε, δ)-definition of a limit line by line (I've never taken a real analysis course).
Let ${\displaystyle f}$ be a real-valued function defined on a subset ${\displaystyle D}$ of the real numbers. Let ${\displaystyle c}$ be a limit point of ${\displaystyle D}$ and let ${\displaystyle L}$ be a real number.
So f is defined on a subset of the real numbers, though an odd subset at that. When I speak of D, it can be the interval [-1,1] or any subinterval including x=0, since they all are pesky.
I have a feeling that the issue here is the condition that c, here 0, has to be a limit point of the subset D of $\mathbb{R}$. To be honest, I only have an intuitive grasp of what a limit point is. I know that the ends of an open interval are its limit points for example, but I have no Idea what it is in general. For a normal space, I know you can just show that a point is a limit point if it is the limit of some sequence of points in the subset at hand. So back to this example, I think you can come up with any other sequence that approaches zero as long as you dodge the gaps.
So, straight to the chase...
$${\displaystyle \lim _{x\to c}f(x)=L\iff (\forall \varepsilon >0,\,\exists \ \delta >0,\,\forall x\in D,\,0<|x-c|<\delta \ \Rightarrow \ |f(x)-L|<\varepsilon )}$$
I'll be honest, I don't fully understand this definition. In fact I've never done an epsilon delta proof before.
If you can respond to some of my reasonings, that'd be great, but any airtight proof/disproof of the existence of the limit is appreciated.
The answer depends on what domain you define $f(x)=\frac{\cos x^{-2}}{\cos x^{-2}}$ on. If it is the whole real line, then the limit does not exist for the reasons you have stated, namely that any interval around 0 (which appears in the epsilon/delta definition: $|x-0|<\delta$) contains infinitely many singularities.
To make this more concrete, suppose I define $f$ such that its value at singularities is 2. Then the limit at 0 still does not exist, because I can find two different sequences $(a_n),(b_n)\to0$ with $\lim_{n\to\infty}f(a_n)=1$ but $\lim_{n\to\infty}f(b_n)=2$.
If, however, you puncture the domain of $f$ so as to exclude the singularities, then use a modified epsilon/delta definition which only considers arguments in the domain of $f$, the limit is 1.
Both these answers agree with Wolfram Alpha.