How do you find the Limit of #ln(1+e^x)/(5x) # as x approaches infinity?

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Answer 1 · 2016-05-21T22:56:08

#1/5#

We can rewrite the argument of the natural logarithm function:

#lim_(xrarroo)ln(1+e^x)/(5x)=lim_(xrarroo)ln(e^x(e^-x+1))/(5x)#

Use the logarithm rule #ln(ab)=ln(a)+ln(b)# to split up the logarithm in the numerator:

#=lim_(xrarroo)(ln(e^x)+ln(e^-x+1))/(5x)=lim_(xrarroo)(x+ln(1/e^x+1))/(5x)#

Split the fraction:

#=lim_(xrarroo)(1/5+ln(1/e^x+1)/(5x))=1/5+lim_(xrarroo)ln(1/e^x+1)/(5x)#

Note that as #x# approaches infinity in this limit, #1/e^oo# goes to #0#:

#=1/5+ln(1/e^oo+1)/(5*oo)=1/5+ln(1)/oo=1/5#

Recall that #ln(1)=0#.