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

[EnumaElish]

[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

$$f'(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)}<br /> {\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}$$

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
$$\frac{-(1-x^2)\pi}<br /> {\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},$$

which is $0/0$ at x=1.

 

[learningphysics]

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...

 

[EnumaElish]

(f'(x))^2 works nicely. Thanks!