Answer: Limit of (1 - (x^.5))/(1 + (x^.5)) as x Approaches Neg. Infinity

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So, I've been mulling over this limit problem for far too long. I feel completely at a loss and refuse to accept my answer of "no limit" or "undefined" regarding the following:

The limit of (1- (x^.5))/(1+(x^.5)) as x approaches NEGATIVE infinity.

Someone care to elaborate on what is actually going on here? As I see it, even multiplying by the conjugate, I still end up with an x variable under a radical and so, the square root of any negative number is imaginary. Can such a number "approaching negative infinity" or "x" can also be a positive number, but moving in the negative direction? Being that as it may, I'm trying to keep the answer within the context of real numbers. So, what is the answer?

-Val-
 
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You can't keep x^(0.5) for x negative in the context of the real numbers. Face it. Once you gotten over that, it certainly has a limit in the complex numbers, regardless of branch choice. The answer is -1.
 
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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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