## Calculus: Early Transcendentals 8th Edition

$$\lim_{t\to0}\frac{e^{2t}-1}{\sin t}=2$$
$$A=\lim_{t\to0}\frac{e^{2t}-1}{\sin t}$$ 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: $$A=\lim_{t\to0}\frac{\frac{d}{dt}(e^{2t}-1)}{\frac{d}{dt}(\sin t)}$$ $$A=\lim_{t\to0}\frac{2e^{2t}}{\cos t}$$ $$A=\frac{2e^{2\times0}}{\cos 0}$$ $$A=\frac{2\times1}{1}=2$$