Is f Continuous Everywhere? Analyzing the Limit of a Fractional Function

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Well, my first question was answered so I figured I would post the second problem I had problems with. It is:

f(x) = lim _{n->\infty}\frac{x^{2n} - 1}{x^{2n} + 1}

Where is f continuous? My first thought is that it is continuous everywhere since I can't find an x value that would make the bottom part of the fraction 0. Isn't that function 1 at for \infty < x \leq-1 or 1 \leq x < \infty and -1 otherwise?
 
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Lol, let's just pretend this question never happened ok?
 
I take it that means you've solved the question then?
 
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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