<s>[INST] Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$ 
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$

My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\dfrac{x}{1-x^2}))+(-\dfrac{x^2}2)}{x^4}$$
$$\lim_{x \to 0} \frac{(\dfrac{(1-\cos (\dfrac{x}{1-x^2}))}{(\dfrac{x}{1-x^2})^2})(\dfrac{x}{1-x^2})^2+(-\dfrac{x^2}2)}{x^4}$$
now what ?
 [/INST] $$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(\frac{x}{1-x^2}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}$$
Now for Taylor's expansion we have that 
$$\lim\limits_{x \to 0}\frac{1}{1-x^2}=1-x^2+o(x^2)$$
So
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x(1-x^2+o(x^2)\right)}{x^4}=$$
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}$$
Now recall that for Taylor's expansion
$$\lim\limits_{x \to 0}\cos t=1-\frac{t^2}{2}+\frac{t^4}{24}+o(t^4)$$
where, in our case, $t=(x-x^3)$. 
Hence
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-1+\frac{x^2}{2}+x^4-\frac{x^4}{24}+o(x^4)}{x^4}=$$
$$=\lim\limits_{x \to 0}\frac{\frac{23}{24}x^4+o(x^4)}{x^4}=\frac{23}{24}.$$
</s>