Question

${lim}_{x\to 0}\frac{e^{2x}}{sin2x}$

Answer 1

We have,

${lim}_{x\to 0}\frac{e^{2x}}{\sin 2x}$

Applying L’ Hospital rule and we get,

${lim}_{x\to 0}\frac{e^{2x}\times 2}{2\cos 2x}$

$={lim}_{x\to 0}\frac{e^{2x}}{\cos 2x}$

Taking limit and we get

$=\frac{e^{2\times 0}}{\cos 2\times 0}$

$=\frac{e^0}{\cos 0}$

$=\frac{1}{1}=1$

Hence, this is the answer.