How do you find $lim (5x+6)/(x^2-4)$ as $x->oo$ ?

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Answer 1 · 2017-01-26T17:58:34

The answer is $=0^-$ and $0^+$

$lim_(x->-oo)(5x+6)/(x^2-4)=lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))$

$lim_(x->-oo)6/x=0$

$lim_(x->-oo)4/x^2=0$

$lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->-oo)5/x=0^-$

$lim_(x->+oo)(5x+6)/(x^2-4)=lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))$

$lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->+oo)5/x=0^+$

How do you find lim (5x+6)/(x^2-4)lim (5x+6)/(x^2-4) as x->oox->oo?