A calculus problem by rudraksh singh

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin x + \ln \left(\sqrt{1+\sin^2 x}-\sin x\right)}{\sin^3 x} & \small \color{#3D99F6} \text{Let } u = \sin x \\ & = \lim_{u \to 0} \frac {u + \ln \left(\sqrt{1+u^2}-u\right)}{u^3} \\ & = \lim_{u \to 0} \frac {u + \ln \left(\frac {(\sqrt{1+u^2}-u)(\sqrt{1+u^2}+u)}{\sqrt{1+u^2}+u} \right)}{u^3} \\ & = \lim_{u \to 0} \frac {u + \ln \left(\frac 1{\sqrt{1+u^2}+u} \right)}{u^3} \\ & = \lim_{u \to 0} \frac {u - \ln \left({\color{#3D99F6}\sqrt{1+u^2}}+u\right)}{u^3} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{u \to 0} \frac {u - \ln \left({\color{#3D99F6}1+\frac {u^2}2- \frac {u^4}8 + ...} + u \right)}{u^3} \\ & = \lim_{u \to 0} \frac {u - {\color{#3D99F6} \ln \left(1+u+\frac {u^2}2- ... \right)}}{u^3} & \small \color{#3D99F6} \text{By Maclaurin series again} \\ & = \lim_{u \to 0} \frac {u - \color{#3D99F6} \left(\left(u+\frac {u^2}2- ...\right) - \frac 12 \left(u+\frac {u^2}2- ...\right)^2 + \frac 13 \left(u+\frac {u^2}2- ...\right)^3 - ... \right)}{u^3} \\ & = \lim_{u \to 0} \frac {u - \color{#3D99F6} \left(\left(u+\frac {u^2}2- ...\right) - \frac 12 \left(u^2+u^3+ ...\right)^2 + \frac 13 \left(u^3+ ...\right)^3 - ... \right)}{u^3} \\ & =\lim_{u \to 0} \frac {\frac 16 u^3 + O(u^4)}{u^3} =\lim_{u \to 0} \left(\frac 16 + \frac {O(u^4)}{u^3} \right) = \lim_{u \to 0} \left(\frac 16 + O(u) \right) \\ & = \frac 16 \approx \boxed{0.1667} \end{aligned}$