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Answer 1 · 2017-12-23T11:23:27

#Lim_(xrarroo)(ln(x+e^x))/x=1#

#Lim_(xrarroo)(ln(x+e^x))/x=oo/oo#

Using L'Hopitals rule:

#Lim_(xrarroo)(1/(x+e^x)*(1+e^x))/1#

#Lim_(xrarroo)(1+e^x)/(x+e^x)=oo/oo#

Again

#Lim_(xrarroo)(e^x)/(1+e^x)=Lim_(xrarroo)cancel(e^x)/(cancel(e^x)(1/(e^x)+1))#

#=Lim_(xrarroo)1/(1/(e^x)+1)=1/(1/oo+1)=1/(0+1)=1#