Can you solve this ?

Question

lim x->+oo [ln(x)/x]

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Answer 1 · 2018-06-10T17:30:56

#Lim_(x->oo)(lnx/x)=0#

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Using L'Hopitale's rule:

#Lim_(x->oo)(lnx/x)=Lim_(x->oo)((d/dx(lnx))/(d/dx(x)))=Lim_(x->oo)((1/x)/1)=0/1=0#

Answer 2 · 2018-06-10T17:34:58

from l'hopital's rule
let #y=Lim_(x->oo)ln(x)/x=Lim_(x->oo)1/(x.1)#
substitute for #x=oo#
#y=1/oo=0#