jbenes
what is the limit as x>9 of (9x)/sqrt(x)3



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LagrangeSon678
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perhaps you can multiply the top and bottom by the conjugate

dinainjune
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6

Outkast3r09
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perhaps you are on the right track lagrangeson

Outkast3r09
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=]

LagrangeSon678
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or we could factor out the top and bottom, yeah that sounds better

jbenes
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how would i factor the top? there are no exponents

Outkast3r09
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it's 6

saifoo.khan
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_

Outkast3r09
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to check just use l'hopitals but 6 is incorrect you're losing a  somewhere

LagrangeSon678
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i made a mistake sorry

dinainjune
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{(9x)/sqrtx 3} x {sqrtx +3/ sqrt x + 3}
{9sqrt x +27x sqrt x 3x} / {x9}
{ sqrt x (x9) 3 (x9)} / {x9}
(x9) (sqrt x 3) / (x9)
= sqrt x 3
=  sqrt 9 3
= 33 =6

jbenes
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ok thanks for all your help!

joemath314159
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lagrange had the right idea, there was just a sign mistake.