Are there unsolved indeterminate limits? I find the question itself is hard to put precisely. I apologize in advance. A simple version could be: 
Let $\mathcal{F}$ be the set of functions obtained via elementary binary operations (sum, product, power, end their inverses) and composition, including polynomials and sinusoidals ($cos(x)$ and $sin(x)$). For example $f(x)=e^{e^x}+cos(x)$. Basically the functions one encounters in basic calculus. The question is the following:
Are there $f,g\in \mathcal{F}$ such that:
$$\lim_{x\to \infty} f(x)=\infty,$$
$$\lim_{x\to \infty} g(x)=\infty,$$ 
but such that the limit $$\lim_{x\to \infty}\dfrac{f(x)}{g(x)}$$ is unsolved? What I mean by this is that no one in the mathematical community knows if the limit exists, or even if existence is guarranteed, no one knows if the limit is finite on infinit.  
 A: Given access to these elementary functions, it's fairly easy to convert statements about, say, rational approximations to $ \pi $ into the existence of a particular limit.
For example, it's unknown if the irrationality measure of $ \pi $ is equal to $ 2 $ or not (in the sense of measure theory, almost all irrational numbers have irrationality measure $ 2 $). It's easy to see that this statement is equivalent to saying that for any $ \epsilon > 0 $,  $ \sin(n) $ is never within $ O(1/n^{2 + \epsilon}) $ of $ 0 $ for all but finitely many values of $ n $, which in turn is equivalent to saying that
$$ \lim_{n \to \infty} \exp(-n^{2 + \epsilon} |\sin(n)|) $$
is equal to $ 0 $ for all $ \epsilon > 0 $, so for instance if you pick $ \epsilon = 1 $ it's unknown if this limit exists or not. If it exists, it's easy to see it must be equal to $ 0 $. If you don't like absolute values, you can instead formulate an equivalent statement as
$$ \lim_{n \to \infty} \exp(-n^{4 + \epsilon} \sin^2(n)) = 0 $$
This is not exactly in the form you want, but you should be able to put it into that form if you multiply and divide by some appropriate function diverging sufficiently quickly to infinity, say $ \exp(\exp(n)) $. This just obscures the basic idea of the construction, however.
A: There are a finite number of mathematicians.  Each mathematician has thought of only a finite number of functions in $\mathcal F$.
 But there are infinitely many many members of $\mathcal F$.
So there are infinitely many $f/g$ that no mathematician has ever
even thought of, therefore doesn't know the limiting behaviour of.
