Limit regarding $\sin x$ , can you make a right choice?

Note that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}= \lim_{x \to 0} \frac{x}{\sin x}=1$ and by L'hospita;'s Rule, $\displaystyle \lim_{x \to 0} \frac{x-\sin x}{x^3} = \lim_{x \to 0} \frac{1-\cos x}{3x^2} = \lim_{x \to 0} \frac{\sin x}{6x}=\frac{1}{6}$ .

Without using Maclaurin series,

$\begin{aligned} \lim_{x \to 0} \left(\frac{x}{\sin^3 x} - \frac{1}{x^2}\right) & = \lim_{x \to 0} \frac{x^3 - \sin ^3x}{x^2\sin^3 x} \\ & = \lim_{x \to 0} \left(\frac{x - \sin x}{x^3} \times \frac{x^2+x\sin x+\sin^2x}{\sin^2x} \times \frac{x}{\sin x} \right) \\ & =\lim_{x \to 0} \frac{x - \sin x}{x^3} \times \lim_{x \to 0}\left(\left(\frac{x}{\sin x}\right)^2 + \frac{x}{\sin x} +1 \right)\times \lim_{x \to 0}\frac{x}{\sin x} \\ & = \frac{1}{6} \times 3 \times 1 \\ & = \boxed{\frac{1}{2}}\end{aligned}$

$\begin{aligned} L&=\lim_{x \to 0} \left(\frac{x}{\sin^3 x} - \frac{1}{x^2}\right) \\&=\lim_{x\to 0}\frac{x^3-\sin ^3 x}{x^2\sin^3 x} \\&=\lim_{x\to 0}\frac{(x-\sin x)(x^2+x\sin x+\sin^2 x)}{x^5} \\&=\lim_{x\to 0}\frac{\left(\frac 16 x^3+\omicron (x^3)\right)(3x^2)}{x^5} \\&=\lim_{x\to 0}\frac {\frac 12 x^5+\omicron (x^5)}{x^5} \\&=\boxed {\frac 12} \end{aligned}$

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac x{\sin^3x}-\frac 1{x^2}\right) & \small \color{#3D99F6} \text{From }\sin 3x = 3\sin x - 4\sin^3 x \\ & = \lim_{x \to 0} \left(\frac {4x}{\color{#3D99F6}3\sin x - \sin 3x}-\frac 1{x^2}\right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \left(\frac {4x}{\color{#3D99F6}3\left(x - \frac {x^3}{3!} + \frac {x^5}{5!} - \cdots\right) - \left(3x - \frac {(3x)^3}{3!} + \frac {(3x)^5}{5!} - \cdots\right)}-\frac 1{x^2}\right) \\ & = \lim_{x \to 0} \left(\frac {4x}{4x^3- 2x^5 + O(x^7)}-\frac 1{x^2}\right) \\ & = \lim_{x \to 0} \left(\frac 1{x^2- \frac 12x^4 + O(x^6)}-\frac 1{x^2}\right) \\ & = \lim_{x \to 0} \frac {x^2 -\left(x^2- \frac 12x^4 + O(x^6)\right)}{x^2 \left(x^2- \frac 12x^4 + O(x^6)\right)} \\ & = \lim_{x \to 0} \frac {\frac 12x^4 - O(x^6)}{x^4- \frac 12x^6 + O(x^8)} & \small \color{#3D99F6} \text{Divide up and down by }x^4 \\ & = \lim_{x \to 0} \frac {\frac 12 - O(x^2)}{1- \frac 12x^2 + O(x^4)} \\ & = \boxed{\frac 12} \end{aligned}$