How do you find the Limit of #ln [(x^.5) + 5] /(lnx)# as x approaches infinity?

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Answer 1 · 2016-08-15T18:45:27

#1/2#

#lim_(x to oo) ln [(sqrt x) + 5] /(lnx)#

this is #oo/oo# indeterminate so we can use L'Hopital

#=lim_(x to oo) (1/(sqrt x + 5) 1/2 x^(-1/2) )/(1/x)#

#= 1/2 lim_(x to oo) (1/(sqrt x + 5)* x^(1/2) )#

#= 1/2 lim_(x to oo) (1/(1+ 5x^(-1/2)) )#

#= 1/2 (lim_(x to oo) 1)/(lim_(x to oo) 1+ 5/sqrt x) #

#= 1/2 * 1/1 = 1/2#