You have to find first $\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}\frac{1)}{\ln x}=0$.
So there an _asymptotic direction_ with slope $0$. However there is no horizontal asymptote, because it would imply the function has a (finite) limit at infinity.
If we had found a non-zero limit $m$ for $\dfrac{f(x)}{x}$, we would have seeked next the limit of $f(x)-mx$ Then there are two main cases:
* If $\lim_{x\to\infty}f(x)-mx=p$, there is an _oblique asymptote_ , with equation $y=mx+p$.
* If $\lim_{x\to\infty}f(x)-mx=\pm\infty$, there is a _parabolic branch_ in the direction with slope $m$.
For functions which have a _Taylor's expansion_ , setting $t=\frac1x$ and considering a Taylor's expansion of $f(\frac1t)$ in a neighbourhood of $t=0$ may yield directly the equation of the asymptote and the position of the curve w.r.t. its asymptote.