Compute  $\lim\limits_{x \rightarrow 0} \frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$

Compute
$$\lim\limits_{x \rightarrow 0} \frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$

Answer

The idea is to use Taylor series. We have
\[\frac{1}{1-x^2}=1+x^2+x^4+x^6\dots.\]
So 
\[\frac{x}{1-x^2}=x+x^3+x^5+x^7\dots .\] 
Also
\[\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}- \dots,\]
and hence
\[\cos (\frac{x}{1-x^2})=1- \frac{(x+x^3+x^5+x^7\dots)^2}{2}+\frac{(x+x^3+x^5+x^7\dots)^4}{4!}+\dots\]
\[=1-\frac{x^2+2x^4+\dots}{2}+\frac{x^4+\dots}{4!}\]
\[=1-\frac{x^2}{2}+(-1+\frac{1}{4!})x^4+\dots =1-\frac{x^2}{2}-\frac{23}{24}x^4+\dots.\]
Thus
\[1-\frac{1}{2}x^2-\cos (\frac{x}{1-x^2})=\frac{23}{24}x^4+\text{Higher Order Terms}.\]
Therefore
\[\lim_{x\rightarrow 0} \frac{1-\frac{1}{2}x^2-\cos (\frac{x}{1-x^2})}{x^4}=\frac{23}{24}.\]

The answer is accepted.