<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] Letting $u = x/(1 - x)^2$, we can write this as:
$$\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u) + \frac{1}{2}(u^{2} - x^{2})}{u^{4}(1 - x^{2})^{4}}$$
$$= \lim_{u \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u)}{u^{4}} + \lim_{x \rightarrow 0}\frac{u^{2} - x^{2}}{2x^{4}}$$
The first limit is a bit more complicated than the limit it said to use- but it's easy with Taylor series or l'Hospital's rule, and the answer is $-\frac{1}{24}$. The second limit is just a rational function in $x$:
$$\lim_{x \rightarrow 0}\frac{u^2 - x^2}{2x^{4}} = \lim_{x \rightarrow 0} \frac{x^{2} - x^{2}(1 - x^{2})^{2}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{x^{2} - x^{2} + 2x^{4} - x^{6}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{2 - x^{2}}{2(1 - x^{2})^{2}} = 1$$
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