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jbenes

  • 4 years ago

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

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  1. LagrangeSon678
    • 4 years ago
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    perhaps you can multiply the top and bottom by the conjugate

  2. dinainjune
    • 4 years ago
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    -6

  3. Outkast3r09
    • 4 years ago
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    perhaps you are on the right track lagrangeson

  4. Outkast3r09
    • 4 years ago
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    =]

  5. LagrangeSon678
    • 4 years ago
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    or we could factor out the top and bottom, yeah that sounds better

  6. jbenes
    • 4 years ago
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    how would i factor the top? there are no exponents

  7. Outkast3r09
    • 4 years ago
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    it's -6

  8. saifoo.khan
    • 4 years ago
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    -_-

  9. Outkast3r09
    • 4 years ago
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    to check just use l'hopitals but 6 is incorrect you're losing a - somewhere

  10. LagrangeSon678
    • 4 years ago
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    i made a mistake sorry

  11. dinainjune
    • 4 years ago
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    {(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

  12. jbenes
    • 4 years ago
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    ok thanks for all your help!

  13. joemath314159
    • 4 years ago
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    lagrange had the right idea, there was just a sign mistake.

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