How to solve the following question without using the l'hospital rule?

Question

#lim_(x to oo) ln(e^x+3x)/x^3#

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Answer 1 · 2018-08-10T21:21:37

#lim_(xto oo) ln(e^x+3x)/x^3=0#

#lim_(xto oo) ln(e^x+3x)/x^3#

#e^x+3x~=_(oo)e^x#, because

#lim_(x to oo)(e^x+3x)/e^x=1+(3x)/e^x=1#

so : #lim_(xto oo) ln(e^x+3x)/x^3#

#=lim_(xtooo)ln(e^x)/x^3#

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

#=0#

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