anonymous
  • anonymous
lim(x->1) (sqrt(x)-1)/(x-1) I know the answer but not how to get to that answer.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1439752558976:dw|
anonymous
  • anonymous
please re post the question correctly
anonymous
  • anonymous
Exactly how do I do that?

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Michele_Laino
  • Michele_Laino
hint, we can write this: \[\Large \frac{{\sqrt x - 1}}{{x - 1}} = \frac{{\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
just watch and learn. if you leave , you lose
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
The main thing I don't understand is how to do that first step (which I assume is conjugation)
Michele_Laino
  • Michele_Laino
no, it is the application of this factorization: \[\Large {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Michele_Laino
  • Michele_Laino
where a^2 = x, and b^2=1
anonymous
  • anonymous
Oh I see it now, thanks for helping it click for me :)
Michele_Laino
  • Michele_Laino
:)

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