calculus - How could I calculate this limit without using L'Hopital's Rule $lim_{xrightarrow0} frac{e^x-1}{sin(2x)}$?

I want to calculate the limit which is above without using L'hopital's rule ;

$$\lim_{x\rightarrow0} \frac{e^x-1}{\sin(2x)}$$

Answer

Using the fact that $$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } =1 } \\ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } =1 } $$ we can conclude that $$\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ \sin { 2x } } } =\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } \frac { 2x }{ \sin { 2x } } } =\frac { 1 }{ 2 } $$