যদি  lim_(xto0){1+xln(1+b^2)}^(1/x)  =  2bsin^2theta  , b>0 এবং  theta in(-pi,pi)  হয় তবে  theta  =? 

Explanation:

Another Explanation (5): ```html bài toán: Cho \(\lim_{x \to 0} (1 + x \ln(1+b^2))^{\frac{1}{x}} = 2b\sin^2\theta\), với \(b > 0\) và \(\theta \in (-\pi, \pi)\). Tìm \(\theta\). Giải: Ta có: \(\lim_{x \to 0} (1 + x \ln(1+b^2))^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{\ln(1 + x \ln(1+b^2))}{x}}\) Áp dụng quy tắc L'Hopital: \(\lim_{x \to 0} \frac{\ln(1 + x \ln(1+b^2))}{x} = \lim_{x \to 0} \frac{\frac{\ln(1+b^2)}{1 + x \ln(1+b^2)}}{1} = \ln(1+b^2)\) Vậy, \(\lim_{x \to 0} (1 + x \ln(1+b^2))^{\frac{1}{x}} = e^{\ln(1+b^2)} = 1+b^2\) Theo đề bài, \(1+b^2 = 2b\sin^2\theta\) \(\sin^2\theta = \frac{1+b^2}{2b}\) Ta có: \((b-1)^2 \ge 0 \Rightarrow b^2 - 2b + 1 \ge 0 \Rightarrow b^2 + 1 \ge 2b \Rightarrow \frac{1+b^2}{2b} \ge 1\) Dấu bằng xảy ra khi \(b=1\). Mà \(\sin^2\theta \le 1\) nên \(\sin^2\theta = 1\) \(\Rightarrow \sin\theta = \pm 1\) \(\theta = \pm \frac{\pi}{2} + k2\pi\) Vì \(\theta \in (-\pi, \pi)\) nên \(\theta = \pm \frac{\pi}{2}\) 🥳 ```