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

Answer 1

Cameron Alexander

#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#

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Answer 2

Adam Adkins

The limit of (x^2 - xln(e^x+1)) as x approaches infinity is infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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