How do you evaluate $(3e^-x+6)/(6e^-x+3)$ as x approaches infinity?

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Answer 1 · 2016-05-16T03:34:18

As $x->+oo$ then $e^-x->0$ hence the limit is

$lim_(x->+oo) (3*e^-x+6)/(6*e^-x+3)=6/3=2$

As $x->-oo$ then $e^-x->oo$ hence the limit is

$lim_(x->-oo) (3*e^-x+6)/(6*e^-x+3)=3/6=1/2$

How do you evaluate (3e^-x+6)/(6e^-x+3)(3e^-x+6)/(6e^-x+3) as x approaches infinity?