How to Compute $\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}$ 
How to compute
$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$
I'm interested in more ways of computing limit for this expression.

My Thoughts
at least i tried to use  L'Hospital's rule but with no luck
\begin{align}
&\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\lim _{x\to \:0}\dfrac{2x\left(\dfrac{\cos \left(x^2\right)}{\sin \left(x^2\right)+1}-1\right)}{\dfrac{2\arcsin \left(x\right)}{\sqrt{1-x^2}}-2x}\\
&=\lim _{x\to \:0}\dfrac{2\left(\dfrac{\cos \left(x^2\right)\left(\sin \left(x^2\right)+1\right)-2x^2\left(\cos ^2\left(x^2\right)+\sin ^2\left(x^2\right)+\sin \left(x^2\right)\right)}{\left(\sin \left(x^2\right)+1\right)^2}-1\right)}{2\left(\dfrac{x\arcsin \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{1-x^2}-1\right)}
\end{align}
Note this limit was taken from competition mathematics so i can't go with L'Hopital because each time i use it i got big terms that i have to derivative for the next time

We will use the following results
$$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)}$$ and $$\lim_{x \to 0}\frac{\arcsin x}{x} = \lim_{t \to 0}\frac{t}{\sin t} = 1\text{ (by putting }t = \arcsin x)$$
We can then proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{(\arcsin x)^{2} - x^{2}}\notag\\
&= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{x^{4}}\cdot\frac{x^{4}}{(\arcsin x)^{2} - x^{2}}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - z}{z^{2}}\cdot\lim_{x \to 0}\frac{x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\text{ (putting }z = x^{2})\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z + \sin z - z}{z^{2}}\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{x}{\arcsin x}\cdot\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - z\cdot\frac{z - \sin z}{z^{3}}\right)\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - 0\cdot\frac{1}{6}\right)\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{(\arcsin x)^{3}}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\frac{\sin^{2}z}{z^{2}}\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\lim_{t \to 0}\frac{t}{t + \sin t}\cdot\frac{t^{3}}{t - \sin t}\text{ (putting }t = \arcsin x)\notag\\
&= 6\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\cdot\lim_{t \to 0}\dfrac{1}{1 + \dfrac{\sin t}{t}}\text{ (putting }u = \sin z)\notag\\
&= 3\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\notag\\
&= 3\lim_{u \to 0}\dfrac{\dfrac{1}{1 + u} - 1}{2u}\text{ (via LHR)}\notag\\
&= -\frac{3}{2}\lim_{u \to 0}\frac{1}{1 + u}\notag\\
&= -\frac{3}{2}\notag
\end{align}
This involves 2 applications of LHR. The use of LHR can be replaced with Taylor series also. Note that both LHR and Taylor series are powerful tools to evaluate limits but they should always be used along with algebraic simplification of limit expression (and use of standard limits) otherwise their usage is a bit complicated.
