Examine Limit: Last Question I Promise!

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Last Question I Promise!
It says given the function (sq root x^2-x)/x-x^2 find the intervals in which it is defined and examine the particular limit. lim x-> 1+ (sq root x^2-x)/x-x^2 Well I found the intervals to be, -infinity<x<0 and 1<x<infinity. So to examine the limit tried plugging in 1+, but you get 0, so then I thought ok rearrange the function.
I tried multiplying by the conjugate and got confused, I am not sure that's the right method since you just end up with (x^2-x)/(x-x^2)(sq root x^2-x).. then you can make the numerator -(x^2-x) and cancel it out with the denominator and your left with 1/(sqroot x^2-x) which is still 1/0... Any suggestions whut I am doing wrong? Thanks Again!
 
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That looks right. In this case, the limit is undefined (or infinity).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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