How do you find the limit of #lnx/x# as #x->oo#?

2 Answers
Jan 16, 2017

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

Explanation:

As we have:

#lim_(x->oo) lnx = +oo#
#lim_(x->oo) x = +oo#

The limit:

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

presents itself in the indeterminate form #oo/oo# and we can use l'Hospital's rule:

#lim f(x)/g(x) = lim (f'(x))/(g'(x))#

so:

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

Sep 25, 2017

On request, giving a demonstration without using l'Hospital:

For #x > 1# we have:

#ln x < x#

if we express #x# as #x = (sqrt(x))^2# then:

#ln x = ln (sqrt(x))^2 = 2ln sqrtx < 2sqrtx#

so:

#lnx/x < 2sqrtx/x = 2/sqrtx#

On the other hand for #x > 1# the logarithm is positive so:

#0 < lnx/x < 2/sqrtx#

Then for #x->oo#

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

based on the squeeze theorem.