Hishoops
(1)
\begin{align} \lim_{x \to 0} \left ( \frac{1}{x (\sin x)^3} - \frac{1}{x^4} - \frac{1}{2x^2} \right ) &= \lim_{x \to 0} \left ( \frac{2x^3 - 2 (\sin x)^3 - x^2 (\sin x)^3}{2x^4 (\sin x)^3} \right ) \end{align}

Since $\displaystyle{\lim_{x \to 0} \left [ 2x^3 - 2 (\sin x)^3 - x^2 (\sin x)^3 \right ]= 0}$ and $\displaystyle{\lim_{x \to 0}\left [ 2x^4 (\sin x)^3 \right ]= 0}$, by L'Hopital's rule:

(2)
\begin{align} \lim_{x \to 0} \left ( \frac{1}{x (\sin x)^3} - \frac{1}{x^4} - \frac{1}{2x^2} \right ) &= \lim_{x \to 0} \frac{6x^2 - 6(\sin x)^2 \cos x - [2x (\sin x)^3 + 3x^2 (\sin x)^2 \cos x]}{8x^3 (\sin x)^3 + 6x^4 (\sin x)^2 \cos x} \\ \end{align}
(3)
\begin{align} = \lim_{x \to 0} \frac{12x - 12 \sin x \cos^2x + 6 (\sin x)^3 - 2(\sin x)^3 - 6x (\sin x)^2 \cos x -6x (\sin x)^2 \cos x - 3x^2[2\sin x \cos^2 x - (\sin x)^3]}{24x^2(\sin x)^3 + 24x^3 (\sin x)^2 \cos x + 24x^3 (\sin x)^2 \cos x + 6x^4[2 \sin x \cos^2 x -(\sin x)^3]} \end{align}
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