If lim_{x rightarrow 0} [1 + x ln(1 + b^2)]^{ | vedclass

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(D) We use the standard limit formula $\lim_{x \rightarrow 0} (1 + f(x))^{\frac{1}{g(x)}} = e^{\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}}$.Given the ex

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

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(D) We use the standard limit formula $\lim_{x ightarrow 0} (1 + f(x))^{\frac{1}{g(x)}} = e^{\lim_{x ightarrow 0} \frac{f(x)}{g(x)}}$.Given the expression $\lim_{x ightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}}$, we have $f(x) = x \ln(1 + b^2)$ and $g(x) = x$.Thus, the limit is $e^{\lim_{x ightarrow 0} \frac{x \ln(1 + b^2)}{x}} = e^{\ln(1 + b^2)} = 1 + b^2$.Equating this to the given expression: $1 + b^2 = 2b \sin^2 	heta$.Rearranging for $\sin^2 	heta$, we get $\sin^2 	heta = \frac{1 + b^2}{2b} = \frac{1}{2} (b + \frac{1}{b})$.Since $b > 0$, by the $AM-GM$ inequality, $b + \frac{1}{b} \geq 2$, which implies $\sin^2 	heta = \frac{1}{2} (b + \frac{1}{b}) \geq 1$.Since the maximum value of $\sin^2 	heta$ is $1$, we must have $\sin^2 	heta = 1$.This implies $\sin 	heta = \pm 1$, so $	heta = \pm \frac{\pi}{2}$.

If $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}} = 2b \sin^2 \theta$, where $b > 0$ and $\theta \in (-\pi, \pi]$, then the value of $\theta$ is

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