Mathematics - Limits Question with Solution | TestHub

Here, ${\lim }_{x\to 0}{\left\{1+x\log \left(1+b^2\right)\right\}}^{1/x}$ $$\begin{array}{rl}& \text{ Given, }\left[1^{\infty }\text{ from }\right]\\ & \Rightarrow e^{{\lim }_{x\to 0}\left\{x\log \left(1+b^2\right)\right\}\cdot \frac{1}{x}}\\ & \Rightarrow e^{\log \left(1+b^2\right)}=(1+b{)}^2\\ & \underset{x\to 0}{\lim }{\left\{1+x\log \left(1+b^2\right)\right\}}^{1/x}=2b{\sin }^2\theta \\ & \Rightarrow \left(1+b^2\right)=2b{\sin }^2\theta \\ & \therefore {\sin }^2\theta =\frac{1+b^2}{2b}\dots \text{ (ii) }\end{array}$$ By $AM\ge GM$, $$\frac{b+\frac{1}{b}}{2}\ge {\left(b\cdot \frac{1}{b}\right)}^{1/2}\Rightarrow \frac{b^2+1}{2b}\ge 1$$ From Eqs. (ii) and (iii), we get ${\sin }^2\theta =1$ $$\Rightarrow \theta =\pm \frac{\pi }{2}\text{ as }\theta \in (-\pi ,\pi ]$$