How do you find the Limit of # (ln x)^2 / (3x )# as x approaches infinity?

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Answer 1 · 2016-08-01T09:11:53

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on inspection you can see that the logarithmic numerator is going to grow more slowly than the denominator

but as its #oo/oo# indeterminate , the easy way to show this is to L'Hopital it

#lim_(x to oo) (ln x)^2 / (3x )#

By Lhopital
#= lim_(x to oo) (2 (ln x)*1/x) / (3 )#

#= 2 lim_(x to oo) ( ln x) / (3x )#

which is still #oo/oo# indeterminate so L'Hopital again

#= 2 lim_(x to oo) ( 1/ x) / (3 )#

#= 2 lim_(x to oo) 1 / (3x ) = 0#

Answer 2 · 2016-08-01T09:33:26

#lim_{x->oo}(log_e x)/(sqrt(x))=0#

#(log_e x)^2/(3x) = 1/3((log_e x)/(sqrt(x)))^2#

Now #e^x# is monotonic so

#lim_{x->oo}(log_e x)/(sqrt(x))equiv lim_{x->oo}(e^{log_e x})/e^{sqrt x} = lim_{x->oo}x/e^{sqrt x} equiv lim_{y->oo}y^2/e^y = 0#

so

#lim_{x->oo}(log_e x)/(sqrt(x))=0#