What is the limit of $xe^(1/x) - x$ as x approaches infinity?

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Answer 1 · 2016-10-24T01:09:57

$lim_(x->oo)(xe^(1/x)-x)=1$

$lim_(x->oo)(xe^(1/x)-x) = lim_(x->oo)x(e^(1/x)-1)$

$=lim_(x->oo)(e^(1/x)-1)/(1/x)$

Direct substitution here produces a $0/0$ indeterminate form. Apply L'Hopital's rule.

$=lim_(x->oo)(d/dx(e^(1/x)-1))/(d/dx1/x)$

$=lim_(x->oo)(e^(1/x)(-1/x^2))/(-1/x^2)$

$=lim_(x->oo)e^(1/x)$

$=e^(1/oo)$

$=e^0$

$=1$

What is the limit of xe^(1/x) - xxe^(1/x) - x as x approaches infinity?