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.

Using $$ \sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3 \, x^5}{40} + \mathcal{O}(x^7) $$ then $$\frac{1}{(\sin^{-1}(x))^2} = \frac{1}{x^2} - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6). $$ This gives $$ \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} = - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6) $$ and the limit $$ \lim_{x \to 0} \left( \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} \right) = - \frac{1}{3}.$$
