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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

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  • $\begingroup$ 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? $\endgroup$ – MPW Dec 2 '19 at 0:21
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    $\begingroup$ 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. $\endgroup$ – imranfat Dec 2 '19 at 0:42
  • $\begingroup$ 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. $\endgroup$ – Hans Lundmark Dec 2 '19 at 9:09
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$y=0$ 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.)

See this article for more information.

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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 $0$.

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