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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 :)

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    what exactly is an empty point for you? – 2017-01-08
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    @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 $c \in \Bbb R$ – 2017-01-08

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

$$\lim_{x\to0}\frac1{\ln(x)-1}=0$$

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 $x=e$...

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    it must be $$x$$ tends to $$0^{+}$$ since we have $$x>0$$ – 2017-01-08
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    :-P Thanks for the nit-pick @Dr.SonnhardGraubner – 2017-01-08
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    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 $$ \lim_{x \to e} \frac{ 3 \ln e +4}{\ln e -1}$$ – 2017-01-08
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    @Ozk Yup. You have to determine if it actually goes to $\pm\infty$ to be an asymptote. – 2017-01-08
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    @SimpleArt I've made another change while you wrote a comment, please review it. – 2017-01-08
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    @Ozk Huh? If the denominator tend to $0$ and the numerator tend to a non-zero value, it should be an asymptote most likely. – 2017-01-08
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    @SimpleArt Why is it? can I assume from it that: $$a \in \Bbb R, a \ne 0 \implies \frac{a}{0} = \pm \infty$$ – 2017-01-08
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    @Ozk Hm... see if you can prove it :D Because it should be like that. – 2017-01-08
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    @SimpleArt Haha, I'm only in high school level (tho I want to start learning at a higher level) so till now in elementary and middle school we've been told that something devided by zero in undefined. Now even the teacher seems not sure about that. – 2017-01-08
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    @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" – 2017-01-08
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    @SimpleArt guess it will be more correct to say that $$a \in \Bbb R, a \ne 0 \implies \frac{a}{0^{\pm}} = \pm \infty$$ – 2017-01-08
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    when $0^{\pm} $ represents a small number from the positive direction and a bot nimber from the negative direction of the X axis, respectively. – 2017-01-08
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    @Ozk Haha... bot nimber... and I think that would be the right direction if you wanted to prove it :D – 2017-01-08