Evaluate limit ((x/(1+(sqrtx))-(1/2))/(x-1)

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Evaluate limit ((x/(1+(sqrtx))-(1/2))/(x-1)

Mathematics
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as x approaches 1
1/2 maybe http://www.wolframalpha.com/input/?i=limit{x+to+0}%28%28x%2F%281%2B%28sqrtx%29%29-%281%2F2%29%29%2F%28x-1%29
I have the answer, I just want to know where I'm going wrong in solving it.

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well then :) ..... whatcha got written?
So far, I've only combined the numerator.
Then I get (2x-1-(sqrtx))/(2*(1+(sqrtx)))/(x-1)
Then I get lost
I set it up where you should multiply by the reciprocal
then took the radical outside of the denominator
now I'm stuck
\[\frac{\frac{x}{(1+\sqrt{x})}-\frac{1}{2}}{x-1}\] this the function?
yes
2x -1 - sqrt(x)) ---------------- right? 2 (1+sqrt(x)) (x-1)
yes.
as x-> 1 we need to determine if the top can control it..... does it factor...
yes, it does.
what do we get for the top then?
let me do the quadratic I guess.
(\[(\sqrt{x}+1)(\sqrt{x}-1/2)\]
I don't think that's right though?
2a^2 -a -1 ; a=sqrt(x) (a-2/2)(a+1/2) a = 1, -1/2 sqrt(x) = 1, -1/2; i agree
oh yes it does. :)
(sqrt(x)-1) (2sqrt(x)+1) -------------------- 2 (sqrt(x)+1) (x-1) we are closer i think lol
I still divide by zero.
gotta go over and make sure I aint missed nuthin
you cancel out the 2(sqrtx+1) on top and bottom.
2x - sqrt(x)) -1 ---------------- 2 (1+sqrt(x)) (x-1) (sqrt.x -1)(2sqrt.x +1) -------------------- (sqrt.x +1) (2x-2) (sqrt.x -1)(2sqrt.x +1) ------------------------ 2x sqrt(x) -2sqrt(x) +2x -2 (sqrt.x -1)(2sqrt.x +1) ------------------------ 2(x sqrt(x) -sqrt(x) +x -1) this is my thougths so far; the bottom factors by grouping maybe
eeek.
2[sqrt(x)(x-1) + 1(x-1)] 2 (sqrt(x)+1) (x-1).... somehow this converts to: 2(sqrt(x)-1)(sqrt(x)+1)^2
http://www.wolframalpha.com/input/?i=%282%2B2sqrt%28x%29%29%28x-1%29
woah, ok.
http://www.wolframalpha.com/input/?i=lim{x+to+1}%28%28x%2F%281%2B%28sqrtx%29%29-%281%2F2%29%29%2F%28x-1%29 this shows steps...
i dont know why it cuts the link short in firefox.... but copy paste that into the address bar and you shouldbe able to see how they stepped thru it
Oh well, that's inconsistent with the answer key. But thanks for all your help. :)

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