I've encountered the two functions: $$f(x) = \frac{1}{\ln x -1}$$ $$g(x) = \ln(8x-x^2)$$ I know that in $x=0$ both functions are undefined, but I can't really understand why in $ g(x)$ there is an asymptote in $ x = 0$ while in $f(x)$ there's an empty point in $x = 0$. Will be happy to an explaination, thanks in advance :)

  • $\begingroup$ what exactly is an empty point for you? $\endgroup$ – user190080 Jan 8 '17 at 12:56
  • $\begingroup$ @user190080 a point on the graph where the function is undefined but still has a $y$ value, like $(0,1)$. wheras a vertical asymptote which is a linear function like $$ x = c$$ when $c \in \Bbb R$ $\endgroup$ – Ozk Jan 8 '17 at 14:02

Notice that as $x\to0^+$, we have


Since $\ln(x)$ gets infinitely big, so the fraction gets infinitely small. Thus, that point is not an asymptote, but just an empty point.

It does, however, have asymptotes at $x=e$...

  • $\begingroup$ it must be $$x$$ tends to $$0^{+}$$ since we have $$x>0$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '17 at 12:59
  • $\begingroup$ :-P Thanks for the nit-pick @Dr.SonnhardGraubner $\endgroup$ – Simply Beautiful Art Jan 8 '17 at 13:00
  • $\begingroup$ Ok, so let me see if I got it. for the function $$ y = \frac{ \ln^3 x +4}{\ln x -1}$$ there might be an asynptote in $ x = e$. But I have to affirm if by computing the limit $$ \lim_{x \to e} \frac{ \ln^3 x +4}{\ln x -1}$$ but it seems peculiar because the denominator is tending to 0 $$ \lim_{x \to e} \frac{ 3 \ln e +4}{\ln e -1}$$ $\endgroup$ – Ozk Jan 8 '17 at 13:19
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    $\begingroup$ @Ozk Well, then I guess I'll just say if there is an infinitely small number in the denominator, the fraction becomes infinitely big, provided the numerator does not cancel the "small-ness" $\endgroup$ – Simply Beautiful Art Jan 8 '17 at 13:32
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    $\begingroup$ when $0^{\pm} $ represents a small number from the positive direction and a bot nimber from the negative direction of the X axis, respectively. $\endgroup$ – Ozk Jan 8 '17 at 13:41

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