Question

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

Answer 1

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

$={lim}_{x\to 0}\frac{(e^{2x}-1)-(e^x-1)}{sin2x}$

$={lim}_{x\to 0}\frac{e^{2x}-1}{sin2x}-{lim}_{x\to 0}\frac{e^x-1}{sin2x}$

$=({lim}_{x\to 0}\frac{e^{2x}-1}{2x}\times {lim}_{x\to 0}\frac{2x}{sin2x})\times \frac{1}{2}$

$=({lim}_{x\to 0}\frac{e^{2x}-1}{x}\times {lim}_{x\to 0}\frac{2x}{sin2x})$

$=1-\frac{1}{2}=\frac{1}{2}$