Question

Solve:
$\underset{x\to 0}{lim}[1+xlog(1+b^2){]}^{1/x}=2b{\sin }^2\theta b>0$ Find $\theta =?$

Answer 1

$\begin{array}{c}{lim}_{x\to 0}{\left[1+x\log \left(1+b^2\right)\right]}^{\frac{1}{x}}=2b{\sin }^2\theta \\ e^{{lim}_{x\to 0}\frac{1}{x}.\left[x\log \left(1+b^2\right)\right]}=2b{\sin }^2\theta \\ 1+b^2=2b{\sin }^2\theta \\ b+\frac{1}{b}=2{\sin }^2\theta \\ {\sin }^2\theta \ge 1\\ {\sin }^2\theta =1\\ \theta =\pm \frac{\pi }{2}\end{array}$