Answer
$$\lim_{t\to0}\frac{e^{2t}-1}{\sin t}=2$$
Work Step by Step
$$A=\lim_{t\to0}\frac{e^{2t}-1}{\sin t}$$
The appearance of both the $e$ and $\sin$ functions tells us that we probably should use L'Hospital's Rule here.
Since $\lim_{t\to0}(e^{2t}-1)=e^{2\times0}-1=1-1=0$ and $\lim_{t\to0}\sin t=\sin0=0.$
This limit is an indeterminate form of $\frac{0}{0}$, so according to L'Hospital's Rule:
$$A=\lim_{t\to0}\frac{\frac{d}{dt}(e^{2t}-1)}{\frac{d}{dt}(\sin t)}$$
$$A=\lim_{t\to0}\frac{2e^{2t}}{\cos t}$$
$$A=\frac{2e^{2\times0}}{\cos 0}$$
$$A=\frac{2\times1}{1}=2$$
