\( \lim _{\theta \rightarrow 0}(f(\theta) \cdot \ln \theta)=j \) when \( f(\theta)=4 \sin (\theta)(\cos (\theta)+1) \)

To evaluate the limit: \[ \lim_{\theta \to 0} \left[ f(\theta) \cdot \ln \theta \right] = j \] where: \[ f(\theta) = 4 \sin(\theta) (\cos(\theta) + 1) \] we proceed as follows: 1. **Simplify \( f(\theta) \) for small \( \theta \):** For small \( \theta \), we can use the approximations: \[ \sin(\theta) \approx \theta \quad \text{and} \quad \cos(\theta) \approx 1 - \frac{\theta^2}{2} \] Substituting these into \( f(\theta) \): \[ f(\theta) \approx 4 \theta \left( 1 + \left(1 - \frac{\theta^2}{2}\right) \right) = 4 \theta \left( 2 - \frac{\theta^2}{2} \right) \approx 8 \theta \] So, for very small \( \theta \): \[ f(\theta) \approx 8 \theta \] 2. **Evaluate the limit:** Substitute the approximation into the original limit: \[ j = \lim_{\theta \to 0} 8 \theta \ln \theta \] Observe the behavior of \( \theta \ln \theta \) as \( \theta \to 0^+ \): - \( \theta \to 0 \) - \( \ln \theta \to -\infty \) However, \( \theta \) approaches zero faster than \( \ln \theta \) approaches negative infinity. To formalize this, consider the substitution \( \theta = e^{-x} \) where \( x \to \infty \): \[ \theta \ln \theta = e^{-x} (-x) = -x e^{-x} \to 0 \quad \text{as} \quad x \to \infty \] Therefore: \[ j = \lim_{\theta \to 0} 8 \theta \ln \theta = 0 \] **Conclusion:** \[ j = 0 \]