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

1 Answer
Jul 19, 2017

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)#