Answer
$2$
Work Step by Step
Given: $\lim\limits_{t\to0}\dfrac{e^{2t}-1}{\sin t}$
Here, $\lim_{t\to0}(e^{2t}-1)=e^{2\times0}-1=1-1=0$ and $\lim_{t\to0}\sin t=\sin0=0.$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{t\to0}\dfrac{\dfrac{d}{dt}(e^{2t}-1)}{\dfrac{d}{dt}(\sin t)}=\lim\limits_{t\to0}\dfrac{2e^{2t}}{\cos t}\\=\dfrac{2e^{2\times0}}{\cos 0}\\=2$