# Limit question

1. Aug 13, 2007

### EnumaElish

[SOLVED] Limit question

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution
$$f'(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)} {\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}$$

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
$$\frac{-(1-x^2)\pi} {\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},$$

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

Last edited: Aug 14, 2007
2. Aug 13, 2007

### 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...

3. Aug 13, 2007

### EnumaElish

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