Evaluating $\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$
I have no idea on how to face this limit. Its value at $0$ seems to be $0$, but its limit equals $\frac{23}{24}$. I tried l'Hôpital but the new expressions were as confusing. I think maybe using squeeze theorem would be useful, but I don't know how to apply it in this case.
Any help would be appreciated, thanks in advance.

Add and subtract $\cos x$ in numerator and then proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\notag\\
&= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos x + \cos x - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\notag\\
&= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos x}{x^{4}} + \dfrac{\cos x - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\tag{1}\\
&= -\frac{1}{24} + 2\lim_{x \to 0}\dfrac{\sin\left(\dfrac{2x - x^{3}}{2(1 - x^{2})}\right)\sin\left(\dfrac{x^{3}}{2(1 - x^{2})}\right)}{x^{4}}\tag{2}\\
&= -\frac{1}{24} + 2\lim_{x \to 0}\dfrac{\sin\left(\dfrac{2x - x^{3}}{2(1 - x^{2})}\right)}{\dfrac{2x - x^{3}}{2(1 - x^{2})}}\cdot\dfrac{2 - x^{2}}{2(1 - x^{2})}\cdot\dfrac{\sin\left(\dfrac{x^{3}}{2(1 - x^{2})}\right)}{\dfrac{x^{3}}{2(1 - x^{2})}}\cdot\dfrac{1}{2(1 - x^{2})}\notag\\
&= -\frac{1}{24} + 2\cdot 1\cdot 1\cdot 1\cdot\frac{1}{2}\notag\\
&= \frac{23}{24}\notag
\end{align}
We have used Taylor series expansion for $\cos x$ while moving from step $(1)$ to step $(2)$.
