Question

${lim}_{x\to 0}\left(\frac{e^{2x}-1}{sin\left(4x\right)}\right)$ is equal to

• A. 0
• B. 0.5
• C. 1
• D. 2

Answer 1

The correct option is B 0.5

${lim}_{x\to 0}\left(\frac{e^{2x}-1}{sin\left(4x\right)}\right)\left(\frac{0}{0}\right)\left(Using{L}^{\prime }Hospitalrule\right)={lim}_{x\to 0}\frac{2e^{2x}}{4cos4x}=\frac{2\times 1}{4\times 1}$

${lim}_{x\to 0}\frac{e^{2x}-1}{sin4x}=\frac{1}{2}$