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?
 A: Taylor expand 
$$\frac x{1-x^2}=x+x^3+x^5+O(x^7),\>\>\>\>\>\cos t = 1-\frac{t^2}2+\frac{t^4}{24}+O(t^6)$$ 
to get
$$\cos\frac{x}{1-x^2}=1-\frac{1}2x^2-\frac{23}{24}x^4+O(x^6)$$
Thus,
$$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
$$=\lim _{x\to 0}\frac{1-\frac{x^2}{2}-(1-\frac{x^2}2-\frac{23}{24}x^4+O(x^6))}{x^4} $$
$$=\lim _{x\to 0}\frac{\frac{23}{24}x^4+O(x^6)}{x^4}=\frac{23}{24}$$
A: 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}
A: Let's put $t=x/(1-x^2)$ so that $t\to 0$ with $x$. Moreover $t/x\to 1$. Now the expression under limit can be written as $$\frac{1-\cos t-(t^2/2)}{t^4}\cdot\frac{t^4}{x^4}+\frac{t^2-x^2}{2x^4}\tag{1}$$ One can easily see via Taylor series (or a little algebra combined with L'Hospital's Rule) that the first term tends to $-1/24$. The second term can be written as $$\frac{t+x} {2x}\cdot\frac{t-x}{x^3}$$ Now the first factor tends to $1$ and $$t-x=\frac{x^3}{1-x^2}$$ so that the second factor also tends to $1$. Therefore the second term in $(1)$ tends to $1$ and the desired limit is $(-1/24)+1=23/24$.
In general one should make use of substitutions (like $t=x/(1-x^2)$) so that the expressions become simpler and easier to type/write. This simplifies the problem considerably. 

In case you are interested in limit of $$\frac{1-\cos t-t^2/2}{t^4}$$ just put $t=2u$ to transform the above expression into $$\frac{\sin^2u-u^2}{8u^4}$$ which is easily factored as $$\frac{1}{8}\cdot\frac{\sin u+u} {u} \cdot\frac{\sin u - u} {u^3}$$ and clearly this tends to $$\frac{1}{8}\cdot 2\cdot\left(-\frac{1}{6}\right)=-\frac{1}{24}$$ The limit $(-1/6)$ of last factor can be easily evaluated by Taylor series or a single application of L'Hospital's Rule. 
