What is the limit of $(3x^2+20x)/(4x^2+9)$ as x goes to infinity?

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Answer 1 · 2015-10-10T12:41:40

$lim_(xrarroo)(3x^2+20x)/(4x^2+9)=3/4$

$(3x^2+20x)/(4x^2+9) = (x^2(3+20/x))/(x^2(4+9/x^2))$ for $x != 0$

$= (3+20/x)/(4+9/x^2)$ for $x != 0$ .

As $xrarroo$ , we get $20/x rarr0$ and $9/x^2 rarr 0$ , so we have

$lim_(xrarroo)(3x^2+20x)/(4x^2+9) = lim_(xrarroo)(3+20/x)/(4+9/x^2) =3/4$

Answer 2 · 2015-10-10T12:42:20

$3/4$

$lim_(x->oo) (3x^2+20x)/(4x^2+9) = lim_(x->oo) ((3x^2)/x^2+(20x)/x^2)/((4x^2)/x^2+9/x^2)= lim_(x->oo) (3+20/x)/(4+9/x^2)=3/4$

Note:

$20/x->0$ when $x->oo$

$9/x^2->0$ when $x->oo$

What is the limit of (3x^2+20x)/(4x^2+9)(3x^2+20x)/(4x^2+9) as x goes to infinity?