Discussion Forum

If $\lim_{x \rightarrow0}[1+x \log (1+b^{2})]^{1/x}=2 b\sin^{2} \theta  b>0$ and $\theta\epsilon (-\pi, \pi]$ , then the value of $\theta$ is 

Answer:

Explanation:

 Here,  $\lim_{x \rightarrow0}[1+x \log (1+b^{2})]^{1/x}$

                                                  $[1^{\infty}$ from ]

$\Rightarrow$    $e^{\lim_{x \rightarrow 0} ( x \log [1+b^{2}))}.\frac{1}{x}$

$\Rightarrow$  $e^{\log(1+b^{2})}=(1+b)^{2}$

 Given, 

    $\lim_{x \rightarrow 0}{(1+x \log (1+b^{2}))}^{1/x}=2b \sin^{2} \theta$

$\Rightarrow$   $(1+b^{2})=2b \sin^{2} \theta$

$\therefore$    $\sin^{2} \theta = \frac{1+b^{2}}{2b}$

By AM $\geq$ GM,

$\frac{b+\frac{1}{b}}{2}\geq \left( b.\frac{1}{b}\right)^{1/2} \Rightarrow \frac{b^{2}+1}{2b}\geq 1$...(iii)

 From Eqs.(ii) and (iii) , we get  $\sin^{2} \theta=1$

$\Rightarrow$   $\theta = \pm \frac{\pi}{2}$ as $\theta \epsilon (- \pi, \pi )$