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

If lim_(x→ 0)[1+xln(1+b²)]^(1/x)=2bsin²θ, b>0 and θ∈(-π,π], then the value of θ is $\lim_{x\to 0}\left[1+x\ln(1+b^2)\right]^{1/x} = 2b\sin^2\theta$ $b > 0,\quad \theta \in (-\pi,\pi]$ $\text{To find: } \theta$ $1.\ \text{evaluate the limit } (1^{\infty})$ $2.\ 1 + b^2 = 2b\sin^2\theta$ $3.\ \text{AM-GM} + \text{range of } \sin^2\theta$ $\left[1+x\ln(1+b^2)\right]^{1/x} \to e^{\frac{1}{x}\cdot x\ln(1+b^2)}$ $= e^{\ln(1+b^2)} = 1 + b^2$ $1 + b^2 = 2b\sin^2\theta$ $\sin^2\theta = \dfrac{1+b^2}{2b}$ $\sin^2\theta = \dfrac{1+b^2}{2b} = \dfrac{b + \frac1b}{2}$ $b + \dfrac1b \ge 2 \Rightarrow \dfrac{b + \frac1b}{2} \ge 1$ $\Rightarrow \sin^2\theta \ge 1$ $0 \le \sin^2\theta \le 1 \text{ and } \sin^2\theta \ge 1 \Rightarrow \sin^2\theta = 1$ $b + \dfrac1b = 2 \Rightarrow b = 1$ $\sin^2\theta = 1 \Rightarrow \theta = \pm\dfrac{\pi}{2} \in (-\pi,\pi]$ $\boxed{\theta = \pm\dfrac{\pi}{2}} \quad \text{option (4)}$