1. **Problem Statement:** Determine whether the functions \(f(x) = \frac{\sin^2(2x)}{4x^5}\) (number 8) and \(f(x) = \frac{1 - \cos x}{1 + \sin x}\) (number 10) diverge to positive or negative infinity as \(x \to 0\). 2. **Recall the limit behavior and approximations:** - For small \(x\), \(\sin x \approx x\) and \(\cos x \approx 1 - \frac{x^2}{2}\). - The sign and degree of the numerator and denominator near zero determine the limit. 3. **Analyze function 8:** \[ f(x) = \frac{\sin^2(2x)}{4x^5} \] - Approximate numerator: \(\sin(2x) \approx 2x\), so \(\sin^2(2x) \approx (2x)^2 = 4x^2\). - Substitute approximation: \[ f(x) \approx \frac{4x^2}{4x^5} = \frac{4x^2}{4x^5} = \frac{1}{x^3} \] - As \(x \to 0^+\), \(\frac{1}{x^3} \to +\infty\). - As \(x \to 0^-\), \(\frac{1}{x^3} \to -\infty\) because \(x^3\) is negative for negative \(x\). **Conclusion for 8:** - The function diverges to positive infinity from the right and negative infinity from the left. 4. **Analyze function 10:** \[ f(x) = \frac{1 - \cos x}{1 + \sin x} \] - Approximate numerator: \(1 - \cos x \approx 1 - \left(1 - \frac{x^2}{2}\right) = \frac{x^2}{2}\). - Approximate denominator: \(1 + \sin x \approx 1 + x\). - Substitute approximations: \[ f(x) \approx \frac{\frac{x^2}{2}}{1 + x} \] - As \(x \to 0\), denominator \(1 + x \to 1\), so \[ f(x) \to \frac{x^2}{2} \to 0 \] - Since numerator and denominator are positive near zero, the function approaches zero, not infinity. **Conclusion for 10:** - The function does not diverge to infinity; it approaches zero. **Final answers:** - Function 8 diverges to \(+\infty\) as \(x \to 0^+\) and to \(-\infty\) as \(x \to 0^-\). - Function 10 approaches 0 and does not diverge to infinity.