How do you find the limit of #(e^x - 3*x)^ (1/x) # as x approaches infinity?

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Answer 1 · 2016-07-03T10:32:22

e

#lim_{x to oo} (e^x - 3x)^ (1/x)#

#= lim_{x to oo} exp( ln (e^x - 3x)^ (1/x))#

#= exp (lim_{x to oo} color{red}{ ln (e^x - 3x)/x )}#

applying L'Hopital to term in red as it amounts to #oo / oo# because #e^x > > 3x# as #x to oo#

#= exp (lim_{x to oo} (1/ (e^x - 3x)* (e^x - 3))/1 )#

#= exp (lim_{x to oo} (e^x - 3)/ (e^x - 3x) )#

#= exp (lim_{x to oo} (1 - 3e^{-x})/ (1 - color{green}{3xe^{-x}}) )#

looking at the green term, we can see already that the exponential term is the dominant term but we can apply L'Hopital to see this through

#lim_{x to oo} 3x e^{-x} = lim_{x to oo} (3x)/e^x = lim_{x to oo} 3/e^x = 0#

so the limit is

#exp ( (1 )/ (1 ))#

#= e#