Evaluate the limit #lim_(x->oo) (x^2-xln(e^x+1))#?

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Answer 1 · 2017-04-09T17:05:16

#lim_(x->oo) (x^2-xln(e^x+1)) = 0#

We want:

#lim_(x->oo) (x^2-xln(e^x+1))#

Note that for large #x# (an in fact moderately small #x#) then #e^x+1 ~~ e^x#

Thus for large #x# we have:

# x^2-xln(e^x+1) ~~ x^2-xln(e^x)#
# " " = x^2-x*xln(e) \ \ \ # (Rules of logs)
# " " = x^2-x^2*1 #
# " " = 0 #

Hence,

#lim_(x->oo) (x^2-xln(e^x+1)) = 0#