Question

What are the asymptote(s) and hole(s), if any, of `f(x) = lnx/x`?

Answer 1

The domain of the function is `D=(0,+oo)`
and it's continious in that domain (there are no holes).

We must calculate `lim_{xrarr0^+}f(x)` so let's do it:

`lim_{xrarr0^+}f(x)=lim_{xrarr0^+}lnx/x=lim_{xrarr0^+}1/x*lnx=(+oo)(-oo)=`

`-oo`

Now because `lim_{xrarr0^+}f(x)=-oo`

the line `x=0` is a vertical asymptode of `f(x)`

Now lets calculate `lim_{xrarr+oo}f(x)`

`lim_{xrarr+oo}f(x)=lim_{xrarr+oo}lnx/x`

This limit is `(+oo)/(+oo)` so we can use L'Hôpital's rule, so :

`lim_{xrarr+oo}lnx/x=lim_{xrarr+oo}(1/x)/1=lim_{xrarr+oo}1/x=0`

so the line `y=0` is a horizontal asymptode of `f(x)`