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Find the limit of the given function

Find the value of the following limit.

    \[ \lim_{x \to 0} \frac{\arcsin (2x) - 2 \arcsin x}{x^3}. \]


We apply L’Hopital’s rule twice,

    \begin{align*}  \lim_{x \to 0} \frac{\arcsin (2x) - 2 \arcsin x}{x^3} &= \lim_{x \to 0} \frac{\frac{2}{\sqrt{1-4x^2}} - \frac{2}{\sqrt{1-x^2}}}{3x^2} \\[9pt]  &= \lim_{x \to 0} \frac{ \frac{8x}{(1-4x^2)^{\frac{3}{2}}} - \frac{2x}{(1-x^2)^{\frac{3}{2}}}}{6x} \\[9pt]  &= \lim_{x \to 0} \frac{ \frac{4}{(1-4x^2)^{\frac{3}{2}}} - \frac{1}{(1-x^2)^{\frac{3}{2}}}}{3} \\[9pt]  &= 1. \end{align*}

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