TITLE: How could I calculate this limit without using L'Hopital's Rule $\lim_{x\rightarrow0} \frac{e^x-1}{\sin(2x)}$?
QUESTION [6 upvotes]: I want to calculate the limit which is above without using L'hopital's rule ;
$$\lim_{x\rightarrow0} \frac{e^x-1}{\sin(2x)}$$

REPLY [5 votes]: Equivalents: $\;\mathrm e^x-1\sim_0 x$, $\;\sin 2x\sim_0 2x$, so $\;\dfrac{\mathrm e^x-1}{\sin 2x}\sim_0\dfrac{x}{2x}=\dfrac12.$