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

1 Answer
Andrea S. · Steve M
Jan 16, 2017

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

Explanation:

As we have:

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

The limit:

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

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/sqrt(x) = lim_(x->oo) (d/(dx)lnx)/(d/(dx)sqrtx) = lim_(x->oo) (1/x)/(1/(2sqrt(x))) = lim_(x->oo) (2sqrt(x))/x=lim_(x->oo) 2/sqrt(x) = 0#

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