Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{\sqrt{1+\sin 3x}-1}{\ln (1+\tan 2x)}\\ =& \underset{x\to 0}{\lim }\frac{(1+\sin 3x)-1}{\{\sqrt{1+\sin 3x+1})\ln (1+\tan 2x)}\\ =& \underset{x\to 0}{\lim }\frac{1}{(\sqrt{1+\sin 3x}+1)}\cdot \frac{\sin 3x}{\ln (1+\tan 2x{)}^{\frac{1}{\tan 2x}}\times \frac{1}{\tan 2x}}\end{array}$ $\begin{array}{rl}& =\underset{x\to 0}{\lim }\frac{1}{(\sqrt{1+\sin 3x}+1)}\cdot \left(\frac{\sin 3x}{3x}\right)\left(\frac{2x}{\tan 2x}\right)\\ & \times \frac{3}{\tan 2x}\cdot \frac{1}{\ln (1+\tan 2x{)}^{\frac{1}{\tan 2x}}}\\ & =\left(\frac{1}{1+1}\right)(1)(1)\left(\frac{3}{2}\right)\frac{1}{\ln e}=\frac{3}{4}.\end{array}$