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anonymous

  • one year ago

lim(x->1) (sqrt(x)-1)/(x-1) I know the answer but not how to get to that answer.

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  1. anonymous
    • one year ago
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    |dw:1439752558976:dw|

  2. anonymous
    • one year ago
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    please re post the question correctly

  3. anonymous
    • one year ago
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    Exactly how do I do that?

  4. Michele_Laino
    • one year ago
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    hint, we can write this: \[\Large \frac{{\sqrt x - 1}}{{x - 1}} = \frac{{\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]

  5. anonymous
    • one year ago
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    If I have to complexly program my text to be easily understandable, instead of simply being able to draw it for you, then I think I will take my leave.

  6. anonymous
    • one year ago
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    just watch and learn. if you leave , you lose

  7. Michele_Laino
    • one year ago
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    if we simplify my expression above, we get this: \[\Large \frac{{\sqrt x - 1}}{{x - 1}} = \frac{{\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \frac{1}{{\sqrt x + 1}}\] and the rightmost side is a continuous function at x=1 @Xep

  8. anonymous
    • one year ago
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    The main thing I don't understand is how to do that first step (which I assume is conjugation)

  9. Michele_Laino
    • one year ago
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    no, it is the application of this factorization: \[\Large {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

  10. Michele_Laino
    • one year ago
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    where a^2 = x, and b^2=1

  11. anonymous
    • one year ago
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    Oh I see it now, thanks for helping it click for me :)

  12. Michele_Laino
    • one year ago
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    :)

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spraguer (Moderator)
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