Finding $\lim_{x\to 0}\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}$ How to evaluate the following limit without L'Hôpital's rule
?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$
My attempt:
$$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\left(\frac{x^2}{(\sin^{-1}(x))^2x^2} - \frac1{x^2}\right)\tag2\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot \frac1{x^2} - \frac1{x^2}\tag3\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag4\\& =\left(1\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag5\\& =\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2} \tag6\\& =0\tag7\end{align}$$
But later I realized, $\lim\limits_{x\to a} f(x)\ g(x) = \lim\limits_{x\to a} f(x)\lim\limits_{x\to a} g(x) $ if and only if both  $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{x\to a} g(x)$ are well defined and finite. Thus moving from $(3)$ to $(4)$ is absolutely wrong.

I don't have any more ideas about the problem.

$$\lim_{x \rightarrow 0} \left( \frac{1}{\arcsin(x)^2} - \frac{1}{x^2} \right)  = \lim_{x \rightarrow 0} \left( \frac{x^2 - \arcsin(x)^2}{\arcsin(x)^2 x^2} \right)$$
Taylor series of $ \arcsin(x) = x + x^3/6 + O(x^5) $ at $x \rightarrow 0$, in this way
$$\lim_{x \rightarrow 0} \left( \frac{x^2 - x^2 - x^4/3 - x^6/36}{(x + x^3/6)^2x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3 - x^6/36}{x^4 +  x^6/3 + x^8/36} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3}{x^4} \right) = - \cfrac{1}{3}$$
