Why doesn't $\ln (x)$ have a horizontal asymptote even though its derivative is $1/x$? My understanding is that the derivative gives the gradient of the function at that point.
So for the function $x^2$, its gradient at point $x=10$ is equal to $20$. 
Extrapolating this to $\ln (x)$, I encounter a problem. 
Since $\frac{d}{dx}\ln (x) = 1/x$, the gradient of $\ln (x)$ at $x=100$ is given by $0.01$. What about as $x \to \infty$? Shouldn't the gradient of $\ln(x)$ tend to $0$? 
And if so, why doesn't $\ln(x)$ have a horizontal asymptote?
 A: Yes, the gradient of $\ln x$ "at infinity" (more precisely, in the limit as $x$ becomes increasingly large) is zero.
What this means is that the larger $x$ is, the less of an increase in $\ln x$ you get by a small change to $x$. It doesn't mean there has to be a horizontal asymptote ("adding up infinitely many infinitesimal things doesn't necessarily give you something finite"). Consider for instance the discrete analogue of $\ln x$, the harmonic series:
$$g(n) = \sum_{i=1}^{n} \frac{1}{i}$$
$g(n)$ diverges for $n\to \infty$, even though each term contributes less and less to the total sum as $i$ gets larger.
EDIT:
Yes, but the gradient going to zero is not sufficient to guarantee that the function approaches some horizontal line. Take a look again at the harmonic series: as $i$ increases the change if $g$ from the $\frac{1}{i}$ term becomes smaller and smaller, but $g(n)$ does not approach any line.
A: To help your intuitions: $f(x)=\sqrt{x}$ has no asymptote at $+\infty$, although $f'(x)$ tends to 0.
A: Let's revisit the definition.

We say $y=f(x)$ has a horizontal asymptote at $y=L$ if either $$\displaystyle\lim_{x\to\infty}f(x)=L\quad\text{ or }\quad\displaystyle\lim_{x\to -\infty}f(x)=L.$$

Since $\ln x$ is only defined for $x>0$, we need only consider
$$
\lim_{x\to\infty}\ln x=\infty
$$
so $y=\ln x$ does not have any horizontal asymptote (just as $y=\sqrt{x}$ does not).

PS Also to correct something from the comments: the definition of a horizontal asymptote is not "a horizontal line that the graph of the function gets closer and closer to, but does not touch". See, for example, $y=e^{-x}\sin x$ which has a horizontal asymptote at $y=0$ even though it intersects $y=0$ infinitely many times. This is a very common error that is propagated by teachers for some reason!
Be guided by definitions rather than well-intentioned but often sloppy heuristics. ;-)
A: A straight line is an asymptote of a curve y=f(x), if the perpendicular distance of a point on the curve to the straight line tends to zero as the point goes towards +/- infinity along the curve.  This definition makes it very precise
Padmanabhan
