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.
 A: 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 $\frac{23}{24}$.
A: 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?
A: 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 $(2)$.
