<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] Use the famous formula $1-\cos 2t=2\sin^{2}t$ to transform the expression into $$2\cdot\frac{\sin^{2}t-(x/2)^{2}}{x^{4}}$$ where $t=x/(2(1-x^{2}))$. Next the numerator needs to be split by adding and subtracting $t^2$. It is easy to check that $(t^2-(x/2)^{2})/x^4\to 1/2$ and hence the desired limit is equal to $$1+2\lim_{x\to 0}\frac{\sin^{2}t-t^2}{t^{4}}\cdot(t/x)^{4}$$ Since $t/x\to 1/2$ we can see that he desired limit is equal to $$1+\frac{1}{8}\lim_{t\to 0}\frac{\sin t +t} {t} \cdot\frac{\sin t-t} {t^3}$$ And then the answer is easily seen to be $1+(1/8)(2)(-1/6)=23/24$. Note that the last limit (which evaluates to $-1/6$) necessitates the use of tools like Taylor series or L'Hospital's Rule.

The limit can not be solved just using $\lim\limits_{x\to 0}\dfrac{1-\cos x} {x^{2}}=\dfrac{1}{2}$ and algebraic manipulation. 
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