Is there standard terminology to describe the not-quite-a-limit behavior of ${\tan( \log x) \over x}$ as $x$ approaches infinity? Suppose I want to describe the long term behavior of ${\tan(\log x) \over x}$ as x increases towards positive real infinity.
Now, 
$$\lim_{x \rightarrow \infty}{\tan(\log x) \over x}$$
obviously doesn't exist.  So it would be wrong to say its limit is 0.
But in some very slightly looser sense, the term obviously approaches 0 aside from the very occasional vertical asymptote.  If you were to pick a point at "random" far, far down the number line (I'm being very imprecise here, I know), it would be an $\epsilon$ from 0 with a probability approaching 1 as the random range you were pulling from got larger.  Various summability methods would also make this fact clear.
Is there either standard terminology for getting this idea across, or standard notation for expressing it?
 A: Following Michael Burr's suggestion in the comments, if we let
$$
f(x) = \frac{\tan(\log x)}{x}
$$
then we can define a running measure of the $x$'s for which $|f(x)| > a$ for some fixed $a > 0$:
$$
m_a(x) = \mu\Bigl( \{ t \in \mathbb R : 1 < t < x \text{ and } |f(t)| > a \} \Bigr).
$$

Claim.
  $$
m_a(x) \sim \frac{2}{\pi a} \log x
$$
  as $x \to \infty$.

This gives us a notion of the density of the "bad" intervals (the intervals where $f(x)$ is not small) in the real line. For instance, if we had had
$$
\lim_{x \to \infty} \frac{m_a(x)}{x} = \frac{1}{2}
$$
we would interpret this to mean that the bad intervals take up roughly half of the real line. In our case $\lim_{x \to \infty} m_a(x)/x = 0$, and it tends to zero rather quickly too, so we can interpret this to mean that, proportionally, the bad intervals are pretty insignificant.
Proof sketch. To calculate the width of the intervals where $|f(x)| > a$ we can start by calculating the points at which $\tan(\log x) = \pm ax$. Away from its poles $f(x)$ will be very close to zero for large $x$, so we only need to investigate neighborhoods of the poles.
Following this idea, the method in this answer can be used to show that, for large $x$, the graph of $y = \tan(\log x)$


*

*intersects the graph of $y=ax$ at $x = e^{(2n+1)\pi/2} - \frac{1}{a} + o(1)$ and

*intersects the graph of $y=-ax$ at $x = e^{(2n+1)\pi/2} + \frac{1}{a} + o(1)$


for $n \in \mathbb N$ with $n \to \infty$. We observe that for large $x$ the length of the intervals for which $|f(x)| > a$ approaches $2/a$, and so for
$$
e^{(2n+1)\pi/2} + \frac{1}{a} + \epsilon < x < e^{(2n+3)\pi/2} - \frac{1}{a} - \epsilon \tag{1}
$$
and $n$ large we have
$$
m_a(x) \approx \sum_{k=0}^{n} \frac{2}{a} \approx \frac{2n}{a}. \tag{2}
$$
For $x$ in the range in $(1)$ we have $n = \frac{\log x}{\pi} + O(1)$, and thus $(2)$ becomes
$$
m_a(x) \approx \frac{2}{\pi a} \log x.
$$
The claim follows from this estimate.
A: It's not clear to me how wide the "bad" intervals are.  If they are sufficiently narrow, then we could say the functions $$f_n(x)=\chi_{[n,\infty)}\frac{\tan(\log x)}{x}$$
converge to the constant zero function $f(x)\equiv 0$ globally in measure.  That would mean that the total measure of the "bad" intervals is bounded, and goes to zero as $n\to \infty$ (that is, as we ignore an increasingly larger initial segment of the $\frac{\tan(\log x)}{x}$ bit).
