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jpsmarinho

lim sqrt(x^2-9)/(2x-6) when x goes to -infinity. I use a software and discover that the lim equals to -1/2, but, when I do, II only encounter the result 1/2

  • one year ago
  • one year ago

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  1. shahzadjalbani
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    • one year ago
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  2. jpsmarinho
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    Ah but has square root in the x^2 -9

    • one year ago
  3. jpsmarinho
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    The lim is the following: {(x^2-9)^(1/2)}/(2x-6)

    • one year ago
  4. shubhamsrg
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    well it should be -1/2 only cant be 1/2 as for x tending to -(a very large no.) ,,the expression gets negative,,clearly.. you can justify this as you may 4 = (-2)^2 and send 4 under sqrt sign with - aside ..

    • one year ago
  5. shubhamsrg
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    if you understand what am saying.

    • one year ago
  6. shahzadjalbani
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    For positive infinity it is 1/2 for negative infinity it is -1/2

    • one year ago
  7. shahzadjalbani
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    @jpsmarinho This is answer of your question..

    • one year ago
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  8. jpsmarinho
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    What you did in first step? I don't understand.

    • one year ago
  9. shahzadjalbani
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    squaring both sides

    • one year ago
  10. shahzadjalbani
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    |dw:1338756191105:dw|

    • one year ago
  11. shahzadjalbani
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    @jpsmarinho Now remove the square y and get the radical over whole function .

    • one year ago
  12. jpsmarinho
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    Aah right!!

    • one year ago
  13. jpsmarinho
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    Other doubt: why I can not consider power the numerator and denominator without equal the expression to some variable?

    • one year ago
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