Limit of f'(x) as x-->1 is -4/\pi

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[SOLVED] Limit question

Homework Statement



Show that the limit of f'(x) as x --> 1 is -4/\pi:

Homework Equations



f(x) = 1 - 4 \arccos\left[\frac 1 2 \left(x+\sqrt{2-x^2}\right)\right]/ \pi

The Attempt at a Solution


f&#039;(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)}<br /> {\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}

Both the numerator and the denominator --> 0 as x-->1. I tried l'Hopital's rule. The derivative of the numerator is -4\sqrt 2/(2-x^2)^{3/2}, which evaluates to -4\sqrt 2 at x = 1. To get the stated answer, the derivative of the denominator should be \pi\sqrt 2 at x=1. But it is actually
<br /> \frac{-(1-x^2)\pi}<br /> {\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},<br />

which is 0/0 at x=1. :rolleyes:
 
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L'Hopital seems to go on and on when you try it with f'(x)...

Find the limit of (f'(x))^2 at x=1, then taking the square root of that limit...
 
(f'(x))^2 works nicely. Thanks!
 
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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