Value of Theta Satisfying Limit Equation with Logarithm and Sine

To find the value of $\theta$, we start by evaluating the limit on the left-hand side of the given equation:

$L = \lim_{x \to 0} \left[1 + x \ln(1+b^2)\right]^{\frac{1}{x}}$

This limit is of the indeterminate form $1^\infty$. Using the standard identity $\lim_{x \to 0} (1 + f(x))^{\frac{1}{g(x)}} = e^{\lim_{x \to 0} \frac{f(x)}{g(x)}}$ where $f(x) \to 0$ as $x \to 0$, we have:

$L = e^{\lim_{x \to 0} \frac{x \ln(1+b^2)}{x}} = e^{\ln(1+b^2)} = 1 + b^2$

Substituting this result back into the original equation gives:

$1 + b^2 = 2b \sin^2 \theta$

Since $b > 0$, we can divide both sides by $2b$:

$\sin^2 \theta = \frac{1 + b^2}{2b} = \frac{1}{2} \left(b + \frac{1}{b}\right)$

By the Arithmetic Mean - Geometric Mean (AM-GM) inequality, for any positive real number $b > 0$:

$b + \frac{1}{b} \ge 2\sqrt{b \cdot \frac{1}{b}} = 2$

with equality holding if and only if $b = 1$.

Consequently, we have:

$\sin^2 \theta = \frac{1}{2} \left(b + \frac{1}{b}\right) \ge \frac{1}{2} (2) = 1$

However, for any real value of $\theta$, the sine function satisfies:

$\sin^2 \theta \le 1$

Therefore, the only way the inequality can be satisfied is when:

$\sin^2 \theta = 1 \quad \text{and} \quad b = 1$

Solving $\sin^2 \theta = 1$ gives:

$\sin \theta = \pm 1$

Given that $\theta \in (-\pi, \pi]$, the values of $\theta$ that satisfy this condition are:

$\theta = \pm \frac{\pi}{2}$

Thus, the correct option is D.