Limit of (arcsen(2x)-2arcsen(x))/x^3 as x->0

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Homework Statement


Limit of (arcsen(2x)-2arcsen(x))/x^3 as x->0


Homework Equations


arsen(x)'=1/Sqrt(1-x^2)


The Attempt at a Solution


It's an 0/0 indeterminancy, to solve it, I had to use L'hopital's rule once simplify de expression, multiplying by the conjugate root. Is there an easier way?
 
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As you don't get rid of the trigonometric functions without l'Hospital, I think this ok.

$$\frac{\frac{2}{\sqrt{1-(2x)^2}}-\frac{2}{\sqrt{1-(x)^2}}}{x^2} = 2 \frac{1}{\frac{1}{\sqrt{1-(2x)^2}}+\frac{1}{\sqrt{1-(x)^2}}} \frac{\frac{1}{1-(2x)^2}-\frac{1}{1-(x)^2}}{x^2} \to \frac{2}{2} \frac{\frac{1}{1-(2x)^2}-\frac{1}{1-(x)^2}}{x^2} \to \dots$$
 
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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