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[SOLVED] Limit question
Show that the limit of f'(x) as x --> 1 is -4/\pi:
f(x) = 1 - 4 \arccos\left[\frac 1 2 \left(x+\sqrt{2-x^2}\right)\right]/ \pi
f'(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.
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'(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.

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