Question

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

Answer 1

perhaps you can multiply the top and bottom by the conjugate

Answer 2

-6

Answer 3

perhaps you are on the right track lagrangeson

Answer 4

=]

Answer 5

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

Answer 6

how would i factor the top? there are no exponents

Answer 7

it's -6

Answer 8

-_-

Answer 9

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

Answer 10

i made a mistake sorry

Answer 11

{(9-x)/sqrtx -3} x {sqrtx +3/ sqrt x + 3}
{9sqrt x +27-x sqrt x -3x} / {x-9}
{- sqrt x (x-9) -3 (x-9)} / {x-9}
(x-9) (-sqrt x -3) / (x-9)
= -sqrt x -3
= - sqrt 9 -3
= -3-3 =-6

Answer 12

ok thanks for all your help!

Answer 13

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