How do you find the limit of #ln(ln(x))/x# as x approaches #oo#?

1 Answer
Bdub
Apr 27, 2016

0

Explanation:

#lim_(x->oo) ln(lnx)/x =ln(ln oo)/oo = ln oo/oo =oo/oo#

This is an indeterminate type so we can use l'Hopital's Rule. Therefore, find the derivative of the top and the derivative of the bottom and then find the limit of their quotient as x goes to infinity.
i.e
#lim_(x->oo) (f'(x))/(g'(x))#

#lim_(x->oo) (1/lnx * 1/x)/1 = lim_(x->oo)1/(xlnx) #

#=1/oo = 0#

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