To evaluate the limit

$\lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\sin^2 x} \right),$

let's proceed as follows:

Step-by-Step Solution:

1. Rewrite the terms using a common denominator:

   $\frac{1}{x^2} - \frac{1}{\sin^2 x} = \frac{\sin^2 x - x^2}{x^2 \sin^2 x}.$

2. Recall Taylor series expansion for $\sin x$ around $x = 0$:

   $\sin x = x - \frac{x^3}{6} + O(x^5) \quad \text{as } x \to 0.$

   Squaring this expansion:

   $\sin^2 x = \left( x - \frac{x^3}{6} + O(x^5) \right)^2 = x^2 - \frac{x^4}{3} + O(x^6).$

3. Substitute $\sin^2 x$ into the expression:

   $\sin^2 x - x^2 = \left( x^2 - \frac{x^4}{3} + O(x^6) \right) - x^2 = -\frac{x^4}{3} + O(x^6).$

   Thus, the numerator becomes:

   $\sin^2 x - x^2 \approx -\frac{x^4}{3} \quad \text{as } x \to 0.$

4. Evaluate the denominator:

   $x^2 \sin^2 x \approx x^2 \cdot x^2 = x^4 \quad \text{as } x \to 0.$

5. Combine the results:

   $\frac{\sin^2 x - x^2}{x^2 \sin^2 x} \approx \frac{-\frac{x^4}{3}}{x^4} = -\frac{1}{3}.$

Conclusion:

$\lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\sin^2 x} \right) = -\frac{1}{3}.$

Would you like more detailed steps on the Taylor series or any related concepts?

Here are 5 related questions to deepen your understanding:

1. How is the Taylor series expansion for $\sin x$ derived?
2. Why does the approximation of $\sin x$ work well near $x = 0$?
3. Can we evaluate the same limit using L'Hôpital's Rule?
4. How would this limit change if we replaced $\sin x$ with $\tan x$?
5. What happens if we evaluate $\lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\cos^2 x} \right)$?

Tip: Taylor series expansions are powerful for approximating functions near a point and can simplify complex limit evaluations around that point!