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jpsmarinho

  • 3 years ago

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

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

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

  4. shubhamsrg
    • 3 years ago
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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 ..

  5. shubhamsrg
    • 3 years ago
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    if you understand what am saying.

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

  7. shahzadjalbani
    • 3 years ago
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    @jpsmarinho This is answer of your question..

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

  9. shahzadjalbani
    • 3 years ago
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    squaring both sides

  10. shahzadjalbani
    • 3 years ago
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    |dw:1338756191105:dw|

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

  12. jpsmarinho
    • 3 years ago
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    Aah right!!

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

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