Question

$\underset{x\to 0}{\lim }\frac{e^{2x}-e^x}{\sin 2x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{e^{2x}-e^x}{\sin \left(2x\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{e^x\left(e^x-1\right)}{\sin 2x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{e^x\left(e^x-1\right)}{x}\times \frac{2x}{\sin 2x}\times \frac{1}{2}\right] \\ =e^0\times 1\times \frac{1}{1}\times \frac{1}{2} \\ =\frac{1}{2} \end{gathered}$$