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,

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

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