<s>[INST] Evaluate $\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
An approach I can think about is to expand $\cos$ using taylor series, is there another approach?
 [/INST] This is another way to get the limit. The idea is the same as Quanto's. In fact.
\begin{eqnarray}
&&\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}\\
&=&\lim _{x\to 0}\frac{2\sin^2(\frac{x}{2(1-x^2)})-\frac{x^2}{2}}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{4\sin^2(\frac{x}{2(1-x^2)})-x^2}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{(2\sin(\frac{x}{2(1-x^2)})-x)(2\sin(\frac{x}{2(1-x^2)})+x)}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{2\sin(\frac{x}{2(1-x^2)})-x}{x^3}\cdot\frac{2\sin(\frac{x}{2(1-x^2)})+x}{x}\\
&=&\lim _{x\to 0}\frac{2(\frac{x^3}{2(1-x^2)}-\frac1{3!}(\frac{x}{2(1-x^2)})^3+O(x^5))}{x^3}\\
&=&\frac{23}{24}.
\end{eqnarray}
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