Limits 2

$\begin{aligned} \mathcal L & = \lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{\sin^3 x} = \lim_{x \to 0} \frac{g(x)}{h(x)} \quad \quad \small \color{#3D99F6}{\text{Since } \lim_{x \to 0} \frac{g(x)}{h(x)} = \frac{0}{0} \text{, we can use l'Hôpital's rule.}} \\ & = \lim_{x \to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{3\sin^2 x \cos x} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down once.}} \\ & = \lim_{x \to 0} \frac{-\frac{2x}{(1+x^2)^2} - \frac{x}{(\sqrt{1-x^2})^3}}{6\sin x \cos^2 x - 3\sin^3 x} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down twice.}} \\ & = \lim_{x \to 0} \frac{\frac{6x^2-2}{(1+x^2)^3} - \frac{2x^2+1}{(\sqrt{1-x^2})^5}}{6\cos^3 x - 12 \sin^3 x - 9\sin^2 x \cos x} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down thrice.}} \\ & = \frac{-3}{6} = - \frac{1}{2} = \boxed{-0.5} \end{aligned}$

Relevant wiki: Limits of Functions

Let us denote the expression by $\mathfrak{S}$

$\mathfrak{S} = \lim_{x\to0} \frac{tan^{-1}x-sin^{-1}x}{sin^3x} = \lim_{x\to0} \frac{(x-\frac{x^3}{3}+\frac{x^5}{5} - \cdots) - (x+\frac{x^3}{6} + \frac{3x^5}{40} + \cdots)}{sin^3x}$

$\mathfrak{S} = \lim_{x\to0} \frac{\cancel{x}-\cancel{x} -x^3(\frac{1}{3}+\frac{1}{6}) + \cdots}{sin^3x} = \lim_{x\to0} \frac{\frac{1}{6}+\frac{1}{3} + \cdots}{\frac{sin^3x}{x^3}}$

All the terms when divided by $x^3$ the coefficients of $x^3$ only become x-free .

Using $\lim_{x\to0}\frac{sinx}{x}=0$

$\mathfrak{S} = -\frac{1}{2}=\boxed{-0.5}$