# Asymptote of function

I have a function $$\frac{\ln x}{x}$$ and I wonder, is $$y=0$$ an asymptote? I mean it is kinda strange that graph is in some place is going through that asymptote. I know it meets the criterium of asymptote, but its kinda strange if you understand me. :D

• Not really strange. Behavior on any finite interval, no matter what, has nothing whatsoever to do with asymptotic behavior as $x\to\infty$. A function could even be identically equal to zero for $x<1000000000$ and still have an asymptote $y=0$, right? – MPW Dec 2 '19 at 0:21
• It is perfectly allowed for a function to intersect its own asymptote. Easy example is y=sinx/x which intersects its horizontal asymptote infinite many times. It is a typical high school misconception that students think that a function cannot intersect its horizontal asymptote. But I understand, because horizontal asymptotes are taught after vertical. A function does not intersect a vertical asymptote. If a graph is not a function (parametrics!) then a graph can also intersect a vertical asymptote. – imranfat Dec 2 '19 at 0:42
• If your intuition clashes with a well-established definition (like your “kinda strange” feeling here), you should try to adjust your intuition to match the actual definition better. – Hans Lundmark Dec 2 '19 at 9:09

$$y=0$$ is a horizontal asymptote of the function since $$\lim_{x\to\infty}\frac{\ln x}{x}=0.$$ (You'll need to use l'Hospital's rule to evaluate that limit.)
Yes, $$y=0$$ is a horizontal asymptote, as $$x \to \infty$$.
We know this because, for large values of $$x$$, $$\,\,0<\ln(x)<\sqrt{x}$$, so $$\displaystyle 0<\frac{\ln(x)}{x}<\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$$, which has an asymptote at $$0$$.