The problem involves evaluating the following limit:

$\lim_{x \to 0} \frac{\left(1 + \frac{1}{2}x^2 - \sqrt{1+x^2}\right) \cos(x^2)}{\cos x - e^{-\frac{x^2}{2}}}.$

Let us solve this step by step.

Step 1: Expand $\sqrt{1 + x^2}$ using a Taylor series around $x = 0$:

The expansion of $\sqrt{1+x^2}$ around $x = 0$ is: $\sqrt{1 + x^2} \approx 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + O(x^6).$

So, $1 + \frac{1}{2}x^2 - \sqrt{1+x^2}$ becomes: $1 + \frac{1}{2}x^2 - \left(1 + \frac{1}{2}x^2 - \frac{1}{8}x^4\right) = -\frac{1}{8}x^4 + O(x^6).$

Step 2: Expand $\cos(x^2)$ around $x = 0$:

The Taylor expansion of $\cos(x^2)$ is: $\cos(x^2) \approx 1 - \frac{x^4}{2} + O(x^8).$

Multiplying this with $-\frac{1}{8}x^4$, we get: $\left(1 + \frac{1}{2}x^2 - \sqrt{1+x^2}\right) \cos(x^2) = \left(-\frac{1}{8}x^4\right)\left(1 - \frac{x^4}{2}\right) + O(x^6) = -\frac{1}{8}x^4 + O(x^6).$

Step 3: Expand the denominator $\cos x - e^{-\frac{x^2}{2}}$:

The Taylor expansions of $\cos x$ and $e^{-\frac{x^2}{2}}$ around $x = 0$ are: $\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24},$ $e^{-\frac{x^2}{2}} \approx 1 - \frac{x^2}{2} + \frac{x^4}{8}.$

So, the denominator becomes: $\cos x - e^{-\frac{x^2}{2}} = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{8}\right),$ $\cos x - e^{-\frac{x^2}{2}} = \frac{x^4}{24} - \frac{x^4}{8} = -\frac{x^4}{12}.$

Step 4: Combine numerator and denominator:

The numerator is: $\left(1 + \frac{1}{2}x^2 - \sqrt{1+x^2}\right)\cos(x^2) = -\frac{1}{8}x^4 + O(x^6).$ The denominator is: $\cos x - e^{-\frac{x^2}{2}} = -\frac{1}{12}x^4 + O(x^6).$

The limit becomes: $\lim_{x \to 0} \frac{-\frac{1}{8}x^4}{-\frac{1}{12}x^4} = \frac{\frac{1}{8}}{\frac{1}{12}} = \frac{12}{8} = \frac{3}{2}.$

Final Answer:

$\boxed{\frac{3}{2}}$

Would you like additional details or clarification? Here are some related questions to explore:

1. What is the role of Taylor series in solving limits like this one?
2. How would the result change if the numerator or denominator were slightly modified?
3. Why is it valid to expand functions near $x = 0$ when finding limits?
4. Could L’Hôpital’s rule be applied here instead of Taylor expansions?
5. How do higher-order terms affect the accuracy of the solution?

Tip: When working with limits involving complicated expressions, always check for opportunities to simplify using series expansions or known approximations!