For lim as x -> -∞, how come lxl becomes -x?

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For lim as x --> -∞, how come lxl becomes -x?

Homework Statement



When you take the limit of a rational function as x approaches (-∞), I am dividing the numerator and denominator by the x value with highest exponent. Supposing the highest x value is √(x^2), so I divide the numerator and denominator by lxl. Then how come, in cases of x--> -∞, I am supposed to replace lxl with -x?

Does this question make sense?
 
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Nevermind, it clicked literally 15 seconds after posting the question. No help needed on this particular question anymore :)
 
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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