calculus limit question: another difficult limit problem I have posted previously on a problem in a similar vein here:
Limit evaluation: very tough question, cannot use L'hopitals rule
I believe this problem is very similar, but it has stumped me.
$$\lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{x^4}$$
Really appreciate it if someone has some insight on this.This comes out to be indeterminate if one plugs in zero.
Following the idea from the link above, I tried to recognize this as derivative evaluated at zero of a function, BUT I could not find the function, because I tried to make this all over x, so that means the function I would create would generate a rational type with x^3 on the bottom.
I guess I should also try to look at some trig limit identities as well.
Hope someone out there can see how to navigate this problem.
P

Like what Emanuele said, asymptotic expansions are useful for this kind of limits, and in fact better than L'Hopital (which fails miserably for some limits):
$\frac{x}{1-x^2} \in x + x^3 + O(x^5) \to 0$ as $x \to 0$
[We keep the error term so that at the end we know the error of the final approximation.]
$\cos( \frac{x}{1-x^2} ) \in 1 - \frac{1}{2} ( \frac{x}{1-x^2} )^2 + \frac{1}{24} ( \frac{x}{1-x^2} )^4 + O( ( \frac{x}{1-x^2} )^6 ) \\
 \subset 1 - \frac{1}{2} (x+x^3+O(x^5))^2 + \frac{1}{24} (x+x^3+O(x^5))^4 + O(x^6) \text{ as } x \to 0 \\
 \subset 1 - \frac{1}{2} x^2 - \frac{23}{24} x^4 + O(x^6) \text{ as } x \to 0$
[We can make the substitution into the Taylor expansion only because the input to $\cos$ tends to 0.]
$\frac{ 1 - \frac{1}{2} x^2 - \cos( \frac{x}{1-x^2} ) }{ x^4 } \in \frac{ \frac{23}{24} x^4 + O(x^6) }{ x^4 } = \frac{23}{24} + O(x^2) \to \frac{23}{24}$ as $x \to 0$
But you must make sure you understand the meaning of the Big-O notation and when and why they can be used. To make it more concrete, you can in many cases find explicit constants for bounds instead of using Big-O notation. For example:
$1 - \frac{1}{2} x^2 + \frac{1}{24} x^4 - \frac{1}{720} x^6 \le cos(x) \le 1 - \frac{1}{2} x^2 + \frac{1}{24} x^4$  [obtained by repeated differentiation and Mean-value theorem]
$x + x^3 \le \frac{x}{1-x^2} \le x + x^3 + 2 x^5$ for sufficiently small $x \ge 0$
$x + x^3 \ge \frac{x}{1-x^2} \ge x + x^3 + 2 x^5$ for sufficiently small $x \le 0$
