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[SOLVED] Limit question
Show that the limit of [itex]f'(x)[/itex] as x --> 1 is [itex]-4/\pi[/itex]:
[tex]f(x) = 1 - 4 \arccos\left[\frac 1 2 \left(x+\sqrt{2-x^2}\right)\right]/ \pi[/tex]
[tex]f'(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)}
{\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}[/tex]
Both the numerator and the denominator --> 0 as x-->1. I tried l'Hopital's rule. The derivative of the numerator is [itex]-4\sqrt 2/(2-x^2)^{3/2}[/itex], which evaluates to [itex]-4\sqrt 2[/itex] at x = 1. To get the stated answer, the derivative of the denominator should be [itex]\pi\sqrt 2[/itex] at x=1. But it is actually
[tex]
\frac{-(1-x^2)\pi}
{\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},
[/tex]
which is [itex]0/0[/itex] at x=1.
Homework Statement
Show that the limit of [itex]f'(x)[/itex] as x --> 1 is [itex]-4/\pi[/itex]:
Homework Equations
[tex]f(x) = 1 - 4 \arccos\left[\frac 1 2 \left(x+\sqrt{2-x^2}\right)\right]/ \pi[/tex]
The Attempt at a Solution
[tex]f'(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)}
{\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}[/tex]
Both the numerator and the denominator --> 0 as x-->1. I tried l'Hopital's rule. The derivative of the numerator is [itex]-4\sqrt 2/(2-x^2)^{3/2}[/itex], which evaluates to [itex]-4\sqrt 2[/itex] at x = 1. To get the stated answer, the derivative of the denominator should be [itex]\pi\sqrt 2[/itex] at x=1. But it is actually
[tex]
\frac{-(1-x^2)\pi}
{\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},
[/tex]
which is [itex]0/0[/itex] at x=1.
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