1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \left( \frac{1}{\sin^2 x} - \frac{1}{x^2} \right).$$ 2. **Recall important formulas and rules:** - For small $x$, $\sin x$ can be approximated by its Taylor series: $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ - Squaring this, $$\sin^2 x = \left(x - \frac{x^3}{6} + \cdots \right)^2 = x^2 - \frac{x^4}{3} + \cdots$$ - We will use this expansion to rewrite $\frac{1}{\sin^2 x}$. 3. **Rewrite the expression using the expansion:** $$\frac{1}{\sin^2 x} = \frac{1}{x^2 - \frac{x^4}{3} + \cdots} = \frac{1}{x^2 \left(1 - \frac{x^2}{3} + \cdots \right)} = \frac{1}{x^2} \cdot \frac{1}{1 - \frac{x^2}{3} + \cdots}.$$ 4. **Use the geometric series expansion for the denominator:** For small $x$, $$\frac{1}{1 - y} = 1 + y + y^2 + \cdots$$ where $y = \frac{x^2}{3} + \cdots$. So, $$\frac{1}{1 - \frac{x^2}{3} + \cdots} = 1 + \frac{x^2}{3} + \cdots.$$ 5. **Combine the terms:** $$\frac{1}{\sin^2 x} = \frac{1}{x^2} \left(1 + \frac{x^2}{3} + \cdots \right) = \frac{1}{x^2} + \frac{1}{3} + \cdots.$$ 6. **Substitute back into the original limit:** $$\frac{1}{\sin^2 x} - \frac{1}{x^2} = \left( \frac{1}{x^2} + \frac{1}{3} + \cdots \right) - \frac{1}{x^2} = \frac{1}{3} + \cdots.$$ 7. **Take the limit as $x \to 0$:** All higher order terms vanish, so $$\lim_{x \to 0} \left( \frac{1}{\sin^2 x} - \frac{1}{x^2} \right) = \frac{1}{3}.$$ **Final answer:** $$\boxed{\frac{1}{3}}.$$