<s>[INST] 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.
 [/INST] Preliminary remark. I turned the previous hint into a full answer because it has been downvoted two times. It seems that the little-o notation is not so popular :-(
Use Taylor'expansion. Note that
$$\frac{x}{1-x^2}=x(1+x^2+o(x^2))=x+x^3+o(x^3),$$
and
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$
Then
$$\cos\left(\frac{x}{1-x^2}\right)=
1-\frac{(x+x^3+o(x^3))^2}{2}+\frac{(x+x^3+o(x^3))^4}{4!}+o(x^4)\\
=1-\frac{x^2+2x^4}{2}+\frac{x^4}{24}+o(x^4)
=1-\frac{x^2}{2}-\frac{23x^4}{24}+o(x^4).
$$
Now it should be easy to verify that the desired limit is $\frac{23}{24}$.
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