<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] HINT:
As suggested using Series Expansion,
$$\cos\dfrac x{1-x^2}=1-\dfrac{\left(\dfrac x{1-x^2}\right)^2}{2!}+\dfrac{\left(\dfrac x{1-x^2}\right)^4}{4!}+\text{ the terms containing higher powers of } x$$
$$1-\dfrac{x^2}2-\cos\dfrac x{1-x^2}$$ $$=\dfrac{x^2}{2(1-x^2)^2}-\dfrac{x^2}2-\dfrac{x^4}{24(1-x^2)^4}+\text{ the terms containing higher powers of } x$$
Now $\dfrac{x^2}{2(1-x^2)^2}-\dfrac{x^2}2=\dfrac{x^2\{1-(1-x^2)^2\}}{2(1-x^2)^2}=\dfrac{2x^4-x^6}{2(1-x^2)^2}$
Can you take it from here?
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