F:[a,infinity)->R is continuous with f(x) > 0 for all x in [a,infinity),

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f:[a,infinity)-->R is continuous with f(x) > 0 for all x in [a,infinity),...

Suppose "a" belongs to R, and f:[a,infinity)-->R is continuous with f(x) > 0 for all x in [a,infinity) and limf(x)=1 (as x goes to infinity). Prove that there exists r>0 such that f(x)>r for all x in [a,infinity).
 
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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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